RD Sharma Solutions Class 8 Algebraic Expressions And Identities Exercise 6.4

RD Sharma Class 8 Solutions Chapter 6 Ex 6.4 PDF Free Download

RD Sharma Solutions Class 8 Chapter 6 Exercise 6.4

 Find the following products: (1-15)

Q 1. \(2a^{3}\)(3a + 5b)

SOLUTION:

To find the product, we will use distributive law as follows:

\(2a^{3}\) (3a + 5b)

= \(2a^{3}\) \(\times\) 3a + \(2a^{3}\) \(\times\) 5b

= (2 x 3)(a3 \(\times\) a) + (2 \(\times\) 5)a3b

= \(\left ( 2 \times 3 \right )a^{3 + 1}\)+ \(\left ( 2 \times 5 \right )a^{3}\)b

= \(6a^{4}\) +\(10a^{3}\)b

Thus, the answer is \(6a^{4}\) +\(10a^{3}\)b.

Q 2. -11a(3a + 2b)

SOLUTION:

To find the product, we will use distributive law as follows:

-11a(3a + 2b)

= (-11a) \(\times\) 3a + (-11a) \(\times\) 2b

= (-11 \(\times\) 3) \(\times\) (a \(\times\) a) + (-11 \(\times\) 2) \(\times\) (a \(\times\) b)

= (-33) \(\times\) (a1+1) + (-22) \(\times\) (a \(\times\) b)

= -33a2 – 22ab

Thus, the answer is -33a2 – 22ab.

Q 3. -5a(7a – 2b)

SOLUTION:

To find the product, we will use distributive law as follows:

-5a(7a – 2b)

= (-5a) \(\times\) 7a + (-5a) \(\times\) (-2b)

= (-5 \(\times\) 7) \(\times\) (a \(\times\) a) + (-5 \(\times\) (-2)) \(\times\) (a \(\times\) b)

= (-35) \(\times\) (a1+1) + (10) \(\times\) (a \(\times\) b)

= -35a2 + 10ab

Thus, the answer is -35a2 + 10ab.

Q 4. -11y2(3y + 7)

SOLUTION:

To find the product, we will use distributive law as follows:

—11y2(3y + 7)

= (-11y2) \(\times\) 3y + (-11y2) \(\times\) 7

= (-11 \(\times\) 3)(y2 \(\times\) y) + (-11 \(\times\) 7) \(\times\) (y2)

= (-33)(y2+1) + (-77) \(\times\) (y2)

= \(- 33 y^{3} – 77 y^{2}\)

Thus, the answer is \(- 33 y^{3} – 77 y^{2}\).

Q 5. \(\frac{6x}{5}\left ( x^{3}+y^{3} \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(\frac{6x}{5}\left ( x^{3}+y^{3} \right )\\= \frac{6x}{5} \times x^{3} + \frac{6x}{5} \times y^{3}\\= \frac{6}{5}\times \left ( x\times x^{3} \right ) + \frac{6}{5}\times \left ( x \times y^{3} \right )\\ = \frac{6}{5}\times \left ( x\times x^{1 + 3} \right ) + \frac{6}{5}\times \left ( x \times y^{3} \right )\\= \frac{6x^{4}}{5} + \frac{6xy^{3}}{5}\)

Thus, the answer is \(\frac{6x^{4}}{5} + \frac{6xy^{3}}{5}\).

Q 6. xy(\(x^{3} – y^{3}\))

SOLUTION:

To find the product, we will use the distributive law in the following way:

xy(\(x^{3} – y^{3}\))

= xy \(\times\) \(x^{3}\) –  xy \(\times\) \(y^{3}\)

= (x \(\times\) \(x^{3}\)) \(\times\) y – x \(\times\) (y \(\times\) \(y^{3}\)  )

= \(x^{1 + 3}\)y – x\(y^{1 + 3}\)

= \(x^{4}\)y – x\(y^{4}\)

Thus, the answer is \(x^{4}\)y – x\(y^{4}\).

Q 7. 0.1y(0.1\(x^{5}\) + 0.1y)

SOLUTION:

To find the product, we will use distributive law as follows:

0.1y(0.1\(x^{5}\) + 0.1y)

= (0.ly) (0.1\(x^{5}\)) + (0.ly)(0.ly)

= (0.1 \(\times\) 0.1)(y \(\times\) \(x^{5}\)) + (0.1 \(\times\) 0.1)(y \(\times\) y)

= (0.1 \(\times\) 0.1) (\(x^{5}\) \(\times\) y) + (0.1 \(\times\) 0.1)( \(y^{1 + 1}\))

= 0.01\(x^{5}\)y + 0.01\(y^{2}\)

Thus, the answer is 0.01\(x^{5}\)y + 0.01\(y^{2}\)

Q 8. \(\left ( \frac{-7}{4} a b^{2} c – \frac{6}{25} a^{2} c^{2} \right )\left ( – 50 a^{2} b^{2} c^{2} \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(\left ( \frac{-7}{4} a b^{2} c – \frac{6}{25} a^{2} c^{2} \right )\left ( – 50 a^{2} b^{2} c^{2} \right )\\ = \left \{ \left ( \frac{-7}{4} a b^{2} c \right )\left ( – 50 a^{2} b^{2} c^{2} \right ) \right \} – \left \{ \left ( \frac{6}{25} a^{2} c^{2} \right )\left (- 50 a^{2} b^{2} c^{2} \right ) \right \}\\ = \left \{ \left \{ \frac{-7}{4}\times \left ( -50 \right ) \right \}\left ( a\times a^{2} \right )\left ( b^{2}\times b^{2} \right )\times \left ( c\times c^{2} \right ) \right \}\) \(- \left \{\left ( \frac{6}{25} \right ) \left ( -50 \right ) \left ( a^{2} \times a^{2}\right )\times b^{2}\times \left ( c^{2}\times c^{2} \right )\right \}\)

= \(\frac{175}{2} a^{3} b^{4} c^{3} – \left ( – 12 a^{4} b^{2} c^{4} \right )\)

= \(\frac{175}{2} a^{3} b^{4} c^{3} + 12 a^{4} b^{2} c^{4}\)

Thus, the answer is \(\frac{175}{2} a^{3} b^{4} c^{3} + 12 a^{4} b^{2} c^{4}\).

Q 9. \(-\frac{8}{27} x y z \left ( \frac{3}{2} x y z^{2} – \frac{9}{4} x y^{2} z^{3}\right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(-\frac{8}{27} x y z \left ( \frac{3}{2} x y z^{2} – \frac{9}{4} x y^{2} z^{3}\right )\\ = \left \{ \left ( -\frac{8}{27} x y z \right )\left ( \frac{3}{2} x y z^{2} \right ) \right \} – \left \{ \left ( -\frac{8}{27} x y z \right )\left ( \frac{9}{4} x y^{2} z^{3} \right ) \right \}\)

= \(\left \{ \left ( \frac{-8}{27} \times \frac{3}{2} \right ) \left ( x \times x \right ) \times \left ( y \times y \right ) \times \left ( z \times z^{2} \right )\right \} – \left \{ \left ( \frac{-8}{27} \times \frac{9}{4} \right ) \left ( x \times x \right ) \times \left ( y \times y^{2} \right ) \times \left ( z \times z^{3} \right )\right \}\)

= \(\left \{ \left (-\frac{8}{27} \times \frac{3}{2} \right )\left ( x^{1+1} y^{1+1} z^{1+2} \right ) \right \} – \left \{ \left ( -\frac{8}{27} \times \frac{9}{4} \right )\left ( x^{1+1} y^{1+2} z^{1+3} \right ) \right \}\)

= \(-\frac{4}{9} x^{2} y^{2} z^{3 } + \frac{2}{3} x^{2} y^{3} z^{4}\)

Thus, the answer is \(-\frac{4}{9} x^{2} y^{2} z^{3 } + \frac{2}{3} x^{2} y^{3} z^{4}\)

Q 10. \(-\frac{4}{27} x y z \left ( \frac{9}{2} x^{2} y z – \frac{3}{4} x y z^{2}\right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(-\frac{4}{27} x y z \left ( \frac{9}{2} x^{2} y z – \frac{3}{4} x y z^{2}\right )\)

= \(\left \{ \left ( -\frac{4}{27} x y z \right )\left ( \frac{9}{2} x^{2} y z \right ) \right \} – \left \{ \left ( -\frac{4}{27} x y z \right )\left ( \frac{3}{4} x y z^{2} \right ) \right \}\)

= \(\left \{ \left (-\frac{4}{27} \times \frac{9}{2} \right )\left ( x^{1+2} y^{1+1} z^{1+1} \right ) \right \} – \left \{ \left ( -\frac{4}{27} \times \frac{3}{4} \right )\left ( x^{1+1} y^{1+1} z^{1+2} \right ) \right \}\)

= \(-\frac{2}{3} x^{3} y^{2} z^{2 } + \frac{1}{9} x^{2} y^{2} z^{3}\)

Thus, the answer is \(-\frac{2}{3} x^{3} y^{2} z^{2 } + \frac{1}{9} x^{2} y^{2} z^{3}\)

Q 11. \(1.5x (10 x^{2} y – 100 x y^{2})\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(1.5x (10 x^{2} y – 100 x y^{2})\\= \left ( 1.5 x \times 10 x^{2} y \right ) – \left ( 1.5 x \times 100 x y^{2} \right )\\ = \left ( 15 x^{1+2} y \right ) – \left ( 150 x^{1+1} y^{2} \right ) \\ = 15 x^{3} y – 150 x^{2} y^{2}\)

Thus, the answer is \(15 x^{3} y – 150 x^{2} y^{2}\).

Q 12. \(4.1 x y\left ( 1.1 x – y \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(4.1 x y\left ( 1.1 x – y \right )\\= \left ( 4.1 x y \times 1.1 x \right ) – \left ( 4.1 x y \times y \right )\\= \left \{ \left ( 4.1 \times 1.1 \right ) \times x y \times x \right \} – \left ( 4.1 x y \times y \right )\\= \left ( 4.51 x^{1+1} y \right ) -\left ( 4.1 x y^{1+1} \right )\\= 4.51 x^{2} y – 4.1 x y^{2}\)

Thus, the answer is \(4.51 x^{2} y – 4.1 x y^{2}\)

Q 13. \(250.5 x y \left ( x z + \frac{y}{10} \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(250.5 x y \left ( x z + \frac{y}{10} \right )\\= 250.5 x y \times x z + 250.5 x y \times \frac{y}{10}\\= 250.5 x^{1+1} y z + 25.05 x y^{1+1}\\= 250.5 x^{2} y z + 25.05 x y^{2}\)

Thus, the answer is \(250.5 x^{2} y z + 25.05 x y^{2}\).

Q 14. \(\frac{7}{5} x^{2} y \left ( \frac{3}{5} x y^{2} + \frac{2}{5} x\right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(\frac{7}{5} x^{2} y \left ( \frac{3}{5} x y^{2} + \frac{2}{5} x\right )\\= \frac{7}{5} x^{2} y \times \frac{3}{5} x y^{2} + \frac{7}{5} x^{2} y \times \frac{2}{5} x\\= \frac{21}{25} x^{2+1} y^{1+2} + \frac{14}{25} x^{2+1} y\\= \frac{21}{25} x^{3} y^{3} + \frac{14}{25} x^{3} y\)

Thus, the answer is \(\frac{21}{25} x^{3} y^{3} + \frac{14}{25} x^{3} y\)

Q 15. \(\frac{4}{3} a \left ( a^{2} + b^{2} – 3 c^{2} \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(\frac{4}{3} a \left ( a^{2} + b^{2} – 3 c^{2} \right )\\= \frac{4}{3} a \times a^{2} + \frac{4}{3} a \times b^{2} – \frac{4}{3} a \times 3 c^{2}\\= \frac{4}{3} a^{1+2} + \frac{4}{3} a b^{2} – 4 a c^{2}\\= \frac{4}{3} a^{3} + \frac{4}{3} a b^{2} – 4 a c^{2}\)

Thus, the answer is \(\frac{4}{3} a^{3} + \frac{4}{3} a b^{2} – 4 a c^{2}\).

Q 16. Find the product \(24 x^{2} \left ( 1 – 2 x \right )\) and evaluate its value for x = 3.

SOLUTION:

To find the product, we will use distributive law as follows:

\(24 x^{2} \left ( 1 – 2 x \right )\\= 24 x^{2} \times 1 – 24 x^{2} \times 2 x\\= 24 x^{2} – 48 x^{1+2}\\= 24 x^{2} – 48 x^{3}\)

Substituting x = 3 in the result, we get

\(24 x^{2} – 48 x^{3}\)

= \(24 \left ( 3 \right )^{2} – 48 \left ( 3 \right )^{3}\\= 24 \times 9 – 48 \times 27\\= 216 – 1296\\= -1080\)

Thus, the product is \(24 x^{2} – 48 x^{3}\) and its value for x = 3 is  -1080.

Q 17. Find the product \(-3 y \left ( x y + y^{2} \right )\) and find its value for x = 4 and y = 5.

SOLUTION:

To find the product, we will use distributive law as follows:

\(-3 y \left ( x y + y^{2} \right )\\= -3 y \times x y + \left ( -3 y \right ) \times y^{2}\\= -3 x y^{1+1} – 3 y^{1+2}\\= -3 x y^{2} – 3 y^{3}\)

Substituting x = 4 and y = 5 in the result, we get

\(-3 x y^{2} – 3 y^{3} \\= -3\left ( 4 \right )\left ( 5 \right )^{2} – 3\left ( 5 \right )^{3}\\= -3\left ( 4 \right )\left ( 25 \right ) – 3\left ( 125 \right )\\= -300 – 375\\= -675\)

Thus, the product is \(-3 x y^{2} – 3 y^{3}\), and its value for x = 4 and y = 5 is -675.

Q 18. Multiply \(-\frac{3}{2} x^{2} y^{3}\) by (2x – y) and verify the answer for x = 1 and y = 2.

SOLUTION:

To find the product, we will use distributive law as follows:

\(-\frac{3}{2} x^{2} y^{3} \times \left ( 2x – y \right )\\= \left (-\frac{3}{2} x^{2} y^{3} \times 2 x \right ) – \left ( -\frac{3}{2} x^{2} y^{3} \times y \right )\\= \left ( -3 x^{2+1} y^{3} \right ) – \left ( -\frac{3}{2} x^{2} y^{3+1} \right )\\= -3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}\)

Substituting x = 1 and y = 2 in the result, we get

\(-3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}\\= -3 \left ( 1 \right )^{3} \left ( 2 \right )^{3} + \frac{3}{2} \left ( 1 \right )^{2} \left ( 2 \right )^{4}\\= -3 \times 1 \times 8 + \frac{3}{2} \times 1 \times 16\\= -24 + 24\\= 0\)

Thus, the product is \(-3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}\), its value for x = 1 and y = 2 is 0.

Q 19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.05:

(i) \(15 y^{2}\left ( 2 – 3 x \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(15 y^{2}\left ( 2 – 3 x \right )\)

= \(15 y^{2} \times 2 – 15 y^{2} \times 3 x\)

= \(30 y^{2} – 45 x y^{2}\)

Substituting x = -1 and y = 0.25 in the result, we get:

\(30 y^{2} – 45 x y^{2} \\= 30 \left ( 0.25 \right )^{2} – 45 \left ( -1 \right ) \left ( 0.25 \right )^{2}\\= 30 \times 0.0625 – \left \{ 45 \times \left ( -1 \right ) \times 0.0625\right \}\\= 1.875 – \left ( -2.8125 \right )\\= 1.875 + 2.8125\\= 4.6875\)

(ii) \(-3 x \left ( y^{2} + z^{2} \right )\)        

           

SOLUTION:

To find the product, we will use distributive law as follows:

\(-3 x \left ( y^{2} + z^{2} \right )\\= -3 x \times y^{2} + \left ( -3 x \right ) \times z^{2}\\= -3 x y^{2} – 3 x z^{2}\)

Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get:

\(-3 x y^{2} – 3 x z^{2}\\= -3 \left ( -1 \right ) \left ( 0.25 \right )^{2} – 3 \left ( -1 \right ) \left ( 0.05 \right )^{2}\\= -3 \left ( -1 \right ) \left ( 0.0625 \right ) – 3 \left ( -1 \right ) \left ( 0.0025 \right )\\= 0.1875 + 0.0075\\= 0.195\)

(iii) \(z^{2} \left ( x – y \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(z^{2} \left ( x – y \right )\\= z^{2} \times x – z^{2} \times y\\= xz^{2} – yz^{2}\)

Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get:

\(x z^{2} – y z^{2}\\= \left ( -1 \right ) \left ( 0.05 \right )^{2} – \left ( 0.25 \right ) \left ( 0.05 \right )^{2}\\= \left ( -1 \right )\left ( 0.0025 \right ) – \left ( 0.25 \right ) \left ( 0.0025 \right )\\= -0.0025 – 0.000625\\= -0.003125\)

(iv) \(x z \left ( x^{2} + y^{2} \right )\)

SOLUTION:

To find the product, we will use distributive law as follows:

\(x z \left ( x^{2} + y^{2} \right )\\= x z \times x^{2} + x z \times y^{2}\\= x^{3} z + x y^{2} z\)

Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get:

\(x^{3} z + x y^{2} z\\= \left ( -1 \right )^{3} \left ( 0.05 \right ) + \left ( -1 \right ) \left ( 0.25 \right )^{2} \left ( 0.05 \right )\\= \left ( -1 \right )\left ( 0.05 \right ) + \left ( -1 \right ) \left ( 0.0625 \right ) \left ( 0.05 \right )\\= -0.05 – 0.003125\\= -0.053125\)

Q 20. Simplify:

(i) \(2 x^{2} \left (x^{3} – x \right ) – 3 x \left ( x^{4} + 2 x \right ) – 2 \left ( x^{4} – 3 x^{2} \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(2 x^{2} \left (x^{3} – x \right ) – 3 x \left ( x^{4} + 2 x \right ) – 2 \left ( x^{4} – 3 x^{2} \right )\\= 2 x^{5} – 2 x^{3} – 3 x^{5} – 6 x^{2} – 2 x^{4} + 6 x^{2}\\= 2 x^{5} – 3 x^{5} – 2 x^{4} – 2 x^{3} – 6 x^{2} + 6 x^{2}\\= -x^{5} – 2 x^{4} – 2 x^{3}\)

(ii) \(x^{3} y \left ( x^{2} – 2 x \right ) + 2 x y \left ( x^{3} – x^{4} \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(x^{3} y \left ( x^{2} – 2 x \right ) + 2 x y \left ( x^{3} – x^{4} \right )\\= x^{5} y – 2 x^{4} y + 2 x^{4} y – 2 x^{5} y\\= x^{5} y – 2 x^{5} y – 2 x^{4} y + 2 x^{4} y\\= -x^{5} y\)

(iii) \(3 a ^{2} + 2 \left ( a + 2 \right ) – 3 a \left ( 2 a + 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(3 a ^{2} + 2 \left ( a + 2 \right ) – 3 a \left ( 2 a + 1 \right )\\= 3 a ^{2} + 2 a + 4 – 6 a ^{2} – 3 a\\= 3 a ^{2} – 6 a ^{2} – 3 a + 4\\= – 3 a ^{2} – a + 4\)

(iv) \(x \left ( x + 4 \right ) + 3 x \left ( 2 x^{2} – 1 \right ) + 4 x^{2} + 4\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(x \left ( x + 4 \right ) + 3 x \left ( 2 x^{2} – 1 \right ) + 4 x^{2} + 4\\= x^{2} + 4 x + 6 x^{3} – 3 x + 4 x^{2} + 4\\= x^{2}+ 4 x^{2} + 4 x – 3 x + 6 x^{3} + 4\\= 5 x^{2} + x + 6 x^{3} +4\)

(v) a(b – c) – b(c – a) – c(a – b)

SOLUTION:

To simplify, we will use distributive law as follows:

a(b – c) – b(c – a) – c(a – b)

= ab – ac – bc + ba – ca + cb

= ab + ba – ac – ca – bc – cb

= 0

(vi) a(b – c) + b(c – a) + c(a – b)

SOLUTION:

To simplify, we will use distributive law as follows:

a(b – c) + b(c – a) + c(a – b)

= ab – ac + bc – ba + ca – cb

= ab – ba – ac + ca + bc – cb

= 0

(vii) \(4 a b \left ( a – b \right ) – 6 a^{2}\left ( b – b^{2} \right ) – 3 b^{2} \left ( 2 a^{2} – a\right ) + 2 a b \left ( b – a \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(4 a b \left ( a – b \right ) – 6 a^{2}\left ( b – b^{2} \right ) – 3 b^{2} \left ( 2 a^{2} – a\right ) + 2 a b \left ( b – a \right )\\= 4 a^{2} b – 4 a b^{2} – 6 a^{2} b + 6 a^{2} b^{2} – 6 b^{2} a^{2} + 3 b^{2} a + 2 a b^{2} – 2 a^{2} b\\= 4 a^{2} b – 6 a^{2} b – 2 a^{2} b – 4 a b^{2} + 3 b^{2} a + 2 a b^{2} + 6 a^{2} b^{2} – 6 b^{2} a^{2}\\= -4 a^{2} b + a b^{2}\)

(viii) \(x^{2} \left ( x^{2} + 1 \right ) – x^{3} \left ( x + 1 \right ) – x \left ( x^{3} – x\right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(x^{2} \left ( x^{2} + 1 \right ) – x^{3} \left ( x + 1 \right ) – x \left ( x^{3} – x\right )\\= x^{4} + x^{2} – x^{4} – x^{3} – x^{4} + x^{2}\\= x^{4} – x^{4} – x^{4} – x^{3} + x^{2}+ x^{2}\\= – x^{4} – x^{3} + 2 x^{2}\)

(ix) \(2 a^{2} + 3 a \left ( 1 – 2 a^{3} \right ) + a \left ( a + 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(2 a^{2} + 3 a \left ( 1 – 2 a^{3} \right ) + a \left ( a + 1 \right )\\= 2 a^{2} + 3 a – 6 a^{4} + a^{2} + a\\= 2 a^{2} + a^{2} + 3 a + a – 6 a^{4}\)

(x) \(a^{2} \left ( 2 a – 1 \right ) + 3 a + a^{3} – 8\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(a^{2} \left ( 2 a – 1 \right ) + 3 a + a^{3} – 8\\= 2 a^{3} – a^{2} + 3 a + a^{3} – 8\\= 2 a^{3} + a^{3} – a^{2} + 3 a – 8\\= 3a^{3} – a^{2} + 3 a – 8\)

(xi) \(\frac{3}{2} x^{2} \left ( x^{2} – 1 \right ) + \frac{1}{4}x^{2} \left ( x^{2} + x \right ) – \frac{3}{4} x \left ( x^{3} – 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\frac{3}{2} x^{2} \left ( x^{2} – 1 \right ) + \frac{1}{4}x^{2} \left ( x^{2} + x \right ) – \frac{3}{4} x \left ( x^{3} – 1 \right )\\= \frac{3}{2} x^{4} – \frac{3}{2} x^{2} + \frac{1}{4}x^{4} + \frac{1}{4}x^{3} – \frac{3}{4} x^{4} + \frac{3}{4} x\\= \frac{3}{2} x^{4} + \frac{1}{4}x^{4} – \frac{3}{4} x^{4} + \frac{1}{4}x^{3} – \frac{3}{2} x^{2} + \frac{3}{4} x\\= x^{4} + \frac{1}{4} x^{3} – \frac{3}{2} x^{2} + \frac{3}{4} x\)

(xii) \(a^{2} b \left ( a – b^{2} \right ) + a b^{2}\left ( 4 a b – 2 a^{2} \right ) – a^{3} b\left ( 1 – 2 b \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(a^{2} b \left ( a – b^{2} \right ) + a b^{2}\left ( 4 a b – 2 a^{2} \right ) – a^{3} b\left ( 1 – 2 b \right )\\= a^{3} b – a^{2} b^{3} + 4 a^{2} b^{3} – 2 a^{3} b^{2} – a^{3} b + 2 a^{3} b^{2}\\= a^{3} b – a^{3} b – a^{2} b^{3} + 4 a^{2} b^{3} – 2 a^{3} b^{2} + 2 a^{3} b^{2}\\= 3 a^{2} b^{3}\)

(xiii) \(a^{2} b \left ( a^{3} – a + 1 \right ) – a b \left ( a^{4} – 2 a^{2} + 2a \right ) – b\left ( a^{3} – a^{2} – 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(a^{2} b \left ( a^{3} – a + 1 \right ) – a b \left ( a^{4} – 2 a^{2} + 2a \right ) – b\left ( a^{3} – a^{2} – 1 \right )\\= a^{5} b – a^{3} b + a^{2} b – a^{5} b + 2 a^{3} b – 2 a^{2} b – a^{3} b + a^{2} b + b\\= a^{5} b – a^{5} b- a^{3} b + 2 a^{3} b – 2 a^{2} b + a^{2} b + b \\= b\)

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