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Laws of Exponents

Simplify:i 2 ...

Question

Simplify:
(i) 2x²(x³ − x) − 3x(x⁴ + 2x) − 2(x⁴ − 3x²)
(ii) x³y(x² − 2x) + 2xy(x³ − x⁴)
(iii) 3a² + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x² − 1) + 4x² + 4
(v) a(b − c) − b(c − a) − c(a − b)
(vi) a(b − c) + b(c − a) + c(a − b)
(vii) 4ab(a − b) − 6a²(b − b²) − 3b²(2a² − a) + 2ab(b − a)
(viii) x²(x² + 1) − x³(x + 1) − x(x³ − x)
(ix) 2a² + 3a(1 − 2a³) + a(a + 1)
(x) a²(2a − 1) + 3a + a³ − 8
(xi) $\frac{3}{2} x^{2} (x^{2} - 1) + \frac{1}{4} x^{2} (x^{2} + x) - \frac{3}{4} x (x^{3} - 1)$
(xii) a²b(a − b²⁾ + ab²(4ab − 2a²) − a³b(1 − 2b)
(xiii) a²b(a³ − a + 1) − ab(a⁴ − 2a² + 2a) − b (a³ − a² − 1)

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Solution

(i) To simplify, we will use distributive law as follows:

$2 x^{2} (x^{3} - x) - 3 x (x^{4} + 2 x) - 2 (x^{4} - 3 x^{2}) = 2 x^{5} - 2 x^{3} - 3 x^{5} - 6 x^{2} - 2 x^{4} + 6 x^{2} = 2 x^{5} - 3 x^{5} - 2 x^{4} - 2 x^{3} - 6 x^{2} + 6 x^{2} = - x^{5} - 2 x^{4} - 2 x^{3}$

(ii) To simplify, we will use distributive law as follows:

$x^{3} y (x^{2} - 2 x) + 2 x y (x^{3} - x^{4}) = x^{5} y - 2 x^{4} y + 2 x^{4} y - 2 x^{5} y = x^{5} y - 2 x^{5} y - 2 x^{4} y + 2 x^{4} y = - x^{5} y$

(iii) To simplify, we will use distributive law as follows:

$3 a^{2} + 2 (a + 2) - 3 a (2 a + 1) = 3 a^{2} + 2 a + 4 - 6 a^{2} - 3 a = 3 a^{2} - 6 a^{2} + 2 a - 3 a + 4 = - 3 a^{2} - a + 4$

(iv) To simplify, we will use distributive law as follows:

$x (x + 4) + 3 x (2 x^{2} - 1) + 4 x^{2} + 4 = x^{2} + 4 x + 6 x^{3} - 3 x + 4 x^{2} + 4 = x^{2} + 4 x^{2} + 4 x - 3 x + 6 x^{3} + 4 = 5 x^{2} + x + 6 x^{3} + 4$

(v) To simplify, we will use distributive law as follows:

$a (b - c) - b (c - a) - c (a - b) = a b - a c - b c + b a - c a + c b = a b + b a - a c - c a - b c + c b = 2 a b - 2 a c$

(vi) To simplify, we will use distributive law as follows:

$a (b - c) + b (c - a) + c (a - b) = a b - a c + b c - b a + c a - c b = a b - b a - a c + c a + b c - c b = 0$

(vii) To simplify, we will use distributive law as follows:

$4 a b (a - b) - 6 a^{2} (b - b^{2}) - 3 b^{2} (2 a^{2} - a) + 2 a b (b - a) = 4 a^{2} b - 4 a b^{2} - 6 a^{2} b + 6 a^{2} b^{2} - 6 b^{2} a^{2} + 3 b^{2} a + 2 a b^{2} - 2 a^{2} b = 4 a^{2} b - 6 a^{2} b - 2 a^{2} b - 4 a b^{2} + 3 b^{2} a + 2 a b^{2} + 6 a^{2} b^{2} - 6 b^{2} a^{2} = - 4 a^{2} b + a b^{2}$

(viii) To simplify, we will use distributive law as follows:

$x^{2} (x^{2} + 1) - x^{3} (x + 1) - x (x^{3} - x) = x^{4} + x^{2} - x^{4} - x^{3} - x^{4} + x^{2} = x^{4} - x^{4} - x^{4} - x^{3} + x^{2} + x^{2} = - x^{4} - x^{3} + 2 x^{2}$

(ix) To simplify, we will use distributive law as follows:

$2 a^{2} + 3 a (1 - 2 a^{3}) + a (a + 1) = 2 a^{2} + 3 a - 6 a^{4} + a^{2} + a = 2 a^{2} + a^{2} + 3 a + a - 6 a^{4} = 3 a^{2} + 4 a - 6 a^{4}$

(x) To simplify, we will use distributive law as follows:

$a^{2} (2 a - 1) + 3 a + a^{3} - 8 = 2 a^{3} - a^{2} + 3 a + a^{3} - 8 = 2 a^{3} + a^{3} - a^{2} + 3 a - 8 = 3 a^{3} - a^{2} + 3 a - 8$

(xi) To simplify, we will use distributive law as follows:

$\frac{3}{2} x^{2} (x^{2} - 1) + \frac{1}{4} x^{2} (x^{2} + x) - \frac{3}{4} x (x^{3} - 1) = \frac{3}{2} x^{4} - \frac{3}{2} x^{2} + \frac{1}{4} x^{4} + \frac{1}{4} x^{3} - \frac{3}{4} x^{4} + \frac{3}{4} x = \frac{3}{2} x^{4} + \frac{1}{4} x^{4} - \frac{3}{4} x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x = (\frac{6 + 1 - 3}{4}) x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x = x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x$

(xii) To simplify, we will use distributive law as follows:

$a^{2} b (a - b^{2}) + a b^{2} (4 a b - 2 a^{2}) - a^{3} b (1 - 2 b) = a^{3} b - a^{2} b^{3} + 4 a^{2} b^{3} - 2 a^{3} b^{2} - a^{3} b + 2 a^{3} b^{2} = a^{3} b - a^{3} b - a^{2} b^{3} + 4 a^{2} b^{3} - 2 a^{3} b^{2} + 2 a^{3} b^{2} = 3 a^{2} b^{3}$

(xiii) To simplify, we will use distributive law as follows:

$a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1) = a^{5} b - a^{3} b + a^{2} b - a^{5} b + 2 a^{3} b - 2 a^{2} b - a^{3} b + a^{2} b + b = a^{5} b - a^{5} b - a^{3} b + 2 a^{3} b - a^{3} b + a^{2} b - 2 a^{2} b + a^{2} b + b = b$

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Similar questions

a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1)

a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1)

is simplified, the answer is

x = \frac{1}{a_{1} +} \frac{a}{a_{2} +} \frac{1}{a_{1} +} \frac{1}{a_{2} +} . . . .,

,

y = \frac{1}{2 a_{1} +} \frac{1}{2 a_{2} +} \frac{1}{2 a_{1} +} \frac{1}{2 a_{2} +} . . . . . .,

z = \frac{1}{3 a_{1} +} \frac{1}{3 a_{2} +} \frac{1}{3 a_{1} +} \frac{1}{3 a - 2 +} . . . .,

x (y^{2} - z^{2}) + 2 y (z^{2} - x^{2}) + 3 z (x^{2} - y^{2}) = 0

{(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots)}^{2} + {(a - \frac{1}{2 a -} \frac{1}{2 a -} \dots)}^{2} = 2 a^{2}

(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots) (a - \frac{1}{2 a -} \frac{1}{2 a -} \dots) = a^{2} - \frac{1}{2 a^{2} -} \frac{1}{2 a^{2} -} \dots

a, b, c

denote positive quantities, prove that

a^{2} + b^{2} + c^{2} > b c + c a + a b;

2 (a^{2} + b^{2} + c^{2}) > b c (b + c) + c a (c + a) + a b (a + b)

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