RD Sharma Solutions Class 9 Chapter 4 Ex 4.4
Q1. Find the following products
(a) (3x + 2y)(9x2 – 6xy + 4y2)
(b) (4x – 5y)(16x2 + 20xy + 25y2)
(c) (7p4 + q)(49p8 – 7p4q + q2)
(d) (\(\frac{x}{2}\)
\(\frac{x^{2}}{4}\)
(e) \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
(f) \((3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})\)
(g) \((\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)\)
(h) \((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\)
(i) (1 – x)(1 + x + x^{2})
(j) (1 + x)(1 – x + x^{2})
(k) \((x^{2} – 1)(x^{4} + x^{2} + 1)\)
(l) \((x^{2} + 1)(x^{6} – x^{3} + 1)\)
Sol :
(a) (3x + 2y)(9x2 – 6xy + 4y2)
Given, (3x + 2y)(9x2 – 6xy + 4y2)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
(3x + 2y)(9x2 – 6xy + 4y2) can we written as
=> (3x + 2y)[(3x)2 – (3x)(2y) + (2y)2)]
=> (3x)3 + (2y)3
=> 27x3 + 8y3
Hence, the value of (3x + 2y)(9x2 – 6xy + 4y2) = 27x3 + 8y3
(b) (4x – 5y)(16x2 + 20xy + 25y2)
Given, (4x – 5y)(16x2 + 20xy + 25y2)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
(4x – 5y)(16x2 + 20xy + 25y2) can we written as
=> (4x – 5y)[(4x)2 + (4x)(5y) + (5y)2)]
=> (4x)3 – (5y)3
=> 16x3 – 25y3
Hence, the value of (4x – 5y)(16x2 + 20xy + 25y2) = 16x3 – 25y3
(c) (7p4 + q)(49p8 – 7p4q + q2)
Given, (7p4 + q)(49p8 – 7p4q + q2)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
(7p4 + q)(49p8 – 7p4q + q2) can be written as
=> (7p4 + q)[(7p4)2 – (7p4)(q) + (q)2)]
=> (7p4)3 + (q)3
=> 343p12 + q3
Hence, the value of (7p4 + q)(49p8 – 7p4q + q2) = 343p12 + q3
(d) (\(\frac{x}{2}\)
Sol :
Given, (\(\frac{x}{2}\)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
(\(\frac{x}{2}\)
=> \((\frac{x}{2} + 2y)[(\frac{x}{2})^{2} – \frac{x}{2}(2y) + (2y)^{2}]\)
=> \((\frac{x}{2})^{3} + (2y)^{3}\)
=> \(\frac{x^{3}}{8} + 8y^{3}\)
(e) \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
Sol:
Given, \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
Can be written as,
=> \((\frac{3}{x} – \frac{5}{y})(\frac{3}{x}\)
=> \((\frac{3}{x})^{3} – (\frac{5}{y})^{3}\)
=> \((\frac{27}{x^{3}}) – (\frac{125}{y^{3}})\)
Hence, the value of \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
(f) \((3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})\)
Sol:
Given, \((3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})\)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})\)
=> \((3 + \frac{5}{x})[(3^{2}) – 3(\frac{5}{x}) + (\frac{5}{x})^{2}]\)
=> \((3)^{3} + (\frac{5}{x})^{3}\)
=> 27 + \(\frac{125}{x^{3}}\)
Hence, the value of \((3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})\)
(g) \((\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)\)
Sol:
Given, \((\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)\)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)\)
=> \((\frac{2}{x} + 3x)[(\frac{2}{x})^{2} + (3x)^{2} – (\frac{2}{x})(3x)]\)
=> \((\frac{2}{x})^{3} + (3x)^{3}\)
=> \(\frac{8}{x^{3}} + 9x^{3}\)
Hence, the value of \((\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)\)
(h) \((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\)
Sol:
Given, \((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\)
=> \((\frac{3}{x} – 2x^{2})[(\frac{3}{x})^{2} + (2x^{2})^{2} – (\frac{3}{x})(2x^{2})]\)
=> \((\frac{3}{x} – 2x^{2})[(\frac{9}{x^{2}}) + 4x^{4} – (\frac{3}{x})(2x^{2})]\)
=> \((\frac{3}{x})^{3} – (2x^{2})^{3}\)
=> \(\frac{27}{x^{3}} – 8x^{6}\)
Hence, \((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\)
(i) (1 – x)(1 + x + x^{2})
Sol:
Given, (1 – x)(1 + x + x^{2})
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
(1 – x)(1 + x + x^{2}) can be written as,
=> (1 – x)[(12 + (1)(x)+ x2)]
=> (1)3 – (x)3
=> 1 – x3
Hence, the value of (1 – x)(1 + x + x^{2}) is 1 – x3
(j) (1 + x)(1 – x + x^{2})
Sol:
Given, (1 + x)(1 – x + x^{2})
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
(1 + x)(1 – x + x^{2}) can be written as,
=> (1 + x)[(12 – (1)(x) + x2)]
=> (1)3 + (x)3
=> 1 + x3
Hence, the value of (1 + x)(1 + x – x^{2}) is 1 + x3
(k) \((x^{2} – 1)(x^{4} + x^{2} + 1)\)
Sol:
Given, \((x^{2} – 1)(x^{4} + x^{2} + 1)\)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((x^{2} – 1)(x^{4} + x^{2} + 1)\)
=> (x2 – 1)[(x2)2 – 12 + (x2)(1)]
=> (x2)3 – 13
=> x6 – 1
Hence, \((x^{2} – 1)(x^{4} + x^{2} + 1)\)
(l) \((x^{2} + 1)(x^{6} – x^{3} + 1)\)
Sol:
Given, \((x^{2} + 1)(x^{6} – x^{3} + 1)\)
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((x^{2} + 1)(x^{6} – x^{3} + 1)\)
=> (x3 + 1)[(x3)2 – (x3)(1) + 12]
=> (x3)3 + 13
=> x9 + 1
Hence, the value of \((x^{2} + 1)(x^{6} – x^{3} + 1)\)
Q2. Find x= 3 and y = -1, Find the values of each of the following using in identity:
(a) (9x2 – 4x2)(81y4 + 36x2y2 + 16x4)
(b) \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
(c) \((\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})\)
(d) \((\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})\)
(e) \((\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})\)
Sol:
(a) (9x2 – 4x2)(81y4 + 36x2y2 + 16x4)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
(9x2 – 4x2)(81y4 + 36x2y2 + 16x4) can be written as,
=> (9x2 – 4x2)[(9y2)2 + (9)(4)x2y2 + (4x2)2]
=> (9y2)3 – (4x2)3
=> 729y6 – 64x6
Substitute the value x = 3, y = -1 in 729y6 – 64x6 we get,
=> 729y6 – 64x6
=> 729(-1)6 – 64(3)6
=> 729(1) – 64(729)
=> 729 – 46656
=> -45927
Hence, the product value of (9x2 – 4x2)(81y4 + 36x2y2 + 16x4) = -45927
(b) \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
Sol:
Given,
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
Can be written as,
=> \((\frac{3}{x} – \frac{x}{3})[(\frac{3}{x})^{2} + (\frac{x}{3})^{2} + (\frac{3}{x})(\frac{x}{3})]\)
=> \((\frac{3}{x})^{3} – (\frac{x}{3})^{3}\)
=> \((\frac{27}{x^{3}}) – (\frac{x^{3}}{27})\)
Substitute x = 3 in eq 1
=> \((\frac{27}{3^{3}}) – (\frac{3^{3}}{27})\)
=> \((\frac{27}{27}) – (\frac{27}{27})\)
=> 0
Hence, the value of \((\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})\)
(c) \((\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})\)
Sol:
Given,
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})\)
Can be written as,
=> \((\frac{x}{7} + \frac{y}{3})[(\frac{x}{7})^{2} + (\frac{y}{3})^{2} – (\frac{x}{7})(\frac{y}{3})]\)
=> \((\frac{x}{7})^{3} + (\frac{y}{3})^{3}\)
=> \((\frac{x^{3}}{343}) + (\frac{y^{3}}{27})\)
Substitute x = 3, y = -1 in eq 1
=> \((\frac{3^{3}}{343}) + (\frac{(-1)^{3}}{27})\)
=> \((\frac{27}{343}) – (\frac{1}{27})\)
Taking least common multiple, we get
=> \(\frac{27*27}{343*27} – \frac{1*343}{27*343}\)
=> \(\frac{729}{9261} – \frac{343}{9261}\)
=> \(\frac{729 – 343}{9261}\)
=> \(\frac{386}{9261}\)
Hence, the value of \((\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})\)
(d) \((\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} -+ \frac{xy}{21})\)
Sol:
Given,
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})\)
Can be written as,
=> \((\frac{x}{4} – \frac{y}{3})[(\frac{x}{4})^{2} + (\frac{y}{3})^{2} + (\frac{x}{4})(\frac{y}{3})]\)
=> \((\frac{x}{4})^{3} – (\frac{y}{3})^{3}\)
=> \((\frac{x^{3}}{64}) – (\frac{y^{3}}{27})\)
Substitute x = 3, y = -1 in eq 1
=> \((\frac{3^{3}}{343}) – (\frac{(-1)^{3}}{27})\)
=> \((\frac{27}{64}) + (\frac{1}{27})\)
Taking least common multiple, we get
=> \(\frac{27*27}{64*27} + \frac{1*64}{27*64}\)
=> \(\frac{729}{1728} + \frac{64}{1728}\)
=> \(\frac{729 + 64}{1728}\)
=> \(\frac{793}{9261}\)
Hence, the value of \((\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})\)
(e) \((\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})\)
Sol:
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})\)
=> \((\frac{5}{x} + 5x)[(\frac{5}{x})^{2} + (5x)^{2} – (\frac{5}{x})(5x)]\)
=> \((\frac{5}{x})^{3} + (5x)^{3}\)
=> \(\frac{125}{x^{3}} + 125x^{3}\)
Substitute x = 3, in eq 1
=> \(\frac{125}{3^{3}} + 125(3)^{3}\)
=> \(\frac{125}{27} + 125*27\)
=> \(\frac{125}{27} + 3375\)
Taking least common multiple, we get
=> \(\frac{125}{27} + \frac{3375*27}{27*1}\)
=> \(\frac{125}{27} + \frac{91125}{27}\)
=> \(\frac{125 + 91125}{27}\)
=> \(\frac{91250}{25}\)
Hence, the value of \((\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})\)
Q3. If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2
Sol :
Given, a + b = 10, ab = 16
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
=> a3 + b3 = (a + b)3 – 3ab(a + b)
=> a3 + b3 = (10)3 – 3(16)(10)
=> a3 + b3 = 1000 – 480
=> a3 + b3 = 520
Substitute , a3 + b3 = 520, a + b = 10 in a3 + b3 = (a + b)(a2 + b2 – ab)
a3 + b3 = (a + b)(a2 + b2 – ab)
520 = 10(a2 + b2 – ab)
\(\frac{520}{10}\)
=> (a2 + b2 – ab) = 52
Now, we need to find (a2 + b2 + ab)
Add and subtract 2ab in a2 + b2 + ab
=> a2 + b2 + ab – 2ab + 2ab
=> (a + b)2 – ab
Substitute a + b = 10, ab
=> a2 + b2 + ab = 102 – 16
= 100 – 16
= 84
Hence, the values of (a2 + b2 – ab) = 52 and (a2 + b2 + ab) = 84
Q4. If a + b = 8 and ab = 6, find the value of a3 + b3
Sol:
Given, a + b = 8 and ab = 6
We know that, a3 + b3 = (a + b)3 – 3ab(a + b)
=> a3 + b3 = (a + b)3 – 3ab(a + b)
=> a3 + b3 = (8)3 – 3(6)(8)
=> a3 + b3 = 512 – 144
=> a3 + b3 = 368
Hence, the value of a3 + b3 is 368
Q5. If a – b = 6 and ab = 20, find the value of a3 – b3
Sol:
Given, a – b = 6 and ab = 20
We know that, a3 – b3 = (a – b)3 + 3ab(a – b)
=> a3 – b3 = (a – b)3 + 3ab(a – b)
=> a3 – b3 = (6)3 + 3(20)(6)
=> a3 – b3 = 216 + 360
=> a3 – b3 = 576
Hence, the value of a3 – b3 is 576
Q6. If x = -2 and y = 1, by using an identity find the value of the following:
(a) (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)
(b) \((\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)\)
(c) \((5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})\)
Sol:
Given,
(a) (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
(4y2 – 9x2)(16y4 + 36x2y2 + 81x4) can be written as,
=> (4y2 – 9x2)[(4x)2 + 4y2*9x2 + (9x2)2)
=> (4y2)3 – (9x2)3
=> 64y6 – 729x6 ———- 1
Substitute x = -2 and y = 1 in eq 1
=> 64y6 – 729x6
=> 64(1)6 – 729(-2)6
=> 64 – 729(64)
=> 64(1 – 729)
=> 64(-728)
=> -46592
Hence, the value of (4y2 – 9x2)(16y4 + 36x2y2 + 81x4) is -46592
(b) \((\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)\)
here x = -2
We know that, a3 – b3 = (a – b)(a2 + b2 + ab)
\((\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)\)
=> \((\frac{2}{x} – \frac{x}{2})[(\frac{2}{x})^{2} + (\frac{x}{2})^{2} + (\frac{2}{x})(\frac{x}{2})]\)
=> \((\frac{2}{x})^{3} – (\frac{x}{2})^{3}\)
=> \((\frac{8}{x^{3}}) – (\frac{x^{3}}{8})\)
Substitute x = -2 in eq 1
=> \((\frac{8}{(-2)^{3}}) – (\frac{(-2)^{3}}{8})\)
=> \((\frac{8}{-8}) – (\frac{-8}{8})\)
=> -1 + 1
=> 0
Hence, the value of \((\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)\)
(c) \((5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})\)
Sol:
We know that, a3 + b3 = (a + b)(a2 + b2 – ab)
\((5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})\)
=> (5y + \(\frac{15}{y}\)
=> (5y)3 + (\(\frac{15}{y}\)
=> 125y3 + (\(\frac{3375}{y^{3}}\)
Substitute y = 1 in eq 1
=> 125(1)3 + (\(\frac{3375}{(1)^{2}}\)
=> 125 + 3375
=> 3500
Hence, the value of \((5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})\)
EXERCISE 4.5
Q1. Find the following products:
(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
(b) (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
(c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
(d) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
Sol:
Given,
(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
we know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
so,
(3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) = (3x)3 + (2y)3 + (2z)3 – 3(3x)(2y)(2z)
= 27x3 + 8y3 + 8z3 – 36xyz
Hence, the value of (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) is 27x3 + 8y3 + 8z3 – 36xyz
(b) (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
we know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
so,
(4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx) = (4x)3 + (-3y)3 + (2z)3 -3(4x)(-3y)(2z)
= 64x3 – 27y3+ 8z3 + 72xyz
Hence, the value of (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx) is 64x3 – 27y3+ 8z3 + 72xyz
(c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
we know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
so,
(2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) = (2a)3 + (-3b)3 + (-2c)3 – 3(2a)(-3b)(-2c)
= 8a3 – 27b3 – 8c3 – 36abc
Hence, the value of (c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) is 8a3 – 27b3 – 8c3 – 36abc
(d) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
we know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
so,
(3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz) = (3x)3 + (-4y)3 + (5z)3 -3(3x)(-4y)(5z)
= 27x3 – 64y3 + 125z3 + 180xyz
Hence, the value of (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz) is 27x3 – 64y3 + 125z3 + 180xyz
Q2. If x + y + z = 8 and xy + yz + zx = 20, Find the value of x3 + y3 + z3 – 3xyz
Sol:
given, x + y + z = 8 and xy + yz + zx = 20
We know that,
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
(x + y + z)2 = x2 + y2 + z2 + 2(20)
(x + y + z)2 = x2 + y2 + z2 + 40
82 = x2 + y2 + z2 + 40
64 – 40 = x2 + y2 + z2
x2 + y2 + z2 = 24
we know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 – 3xyz = (x + y + z)[(x2 + y2 + z2) – (xy + yz + zx)]
here, x + y + z = 8, xy + yz + zx = 20, x2 + y2 + z2 = 24
x3 + y3 + z3 – 3xyz = 8[(24 – 20)]
= 8 * 4
= 32
Hence, the value of x3 + y3 + z3 – 3xyz is 32
Q3. If a + b + c = 9 and ab + bc + ca = 26, Find the value of a3 + b3 + c3 – 3abc
Sol :
Given, a + b + c = 9 and ab + bc + ca = 26
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b + c)2 = a2 + b2 + c2 + 2(26)
(a + b + c)2 = a2 + b2 + c2 + 52
92 = a2 + b2 + c2 + 52
81 – 52 = a2 + b2 + c2
a2 + b2 + c2 = 29
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 – 3abc = (a + b + c)[(a2 + b2 + c2) – (ab + bc + ca)]
here, a + b + c = 9, ab + bc + ca = 26, a2 + b2 + c2 = 29
a3 + b3 + c3 – 3abc = 9[(29 – 26)]
= 9 * 3
= 27
Hence, the value of a3 + b3 + c3 – 3abc is 27
Q4. If a + b + c = 9 and a2 + b2 + c2 = 35, Find the value of a3 + b3 + c3 – 3abc
Sol:
Given, a + b + c = 9 and a2 + b2 + c2 = 35
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
92 = 35 + 2(ab + bc + ca)
81 = 35 + 2(ab + bc + ca)
81 – 35 = 2(ab + bc + ca)
\(\frac{46}{2}\)
ab + bc + ca = 23
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 – 3abc = (a + b + c)[(a2 + b2 + c2) – (ab + bc + ca)]
here, a + b + c = 9, ab + bc + ca = 23, a2 + b2 + c2 = 35
a3 + b3 + c3 – 3abc = 9[(35 – 23)]
= 9 * 12
= 108
Hence, the value of a3 + b3 + c3 – 3abc is 108
Q5. Evaluate:
(a) 253 – 753 + 503
(b) 483 – 303 – 183
(c) \((\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}\)
(d) (0.2)3 – (0.3)3 + (0.1)3
Sol:
Given,
(a) 253 – 753 + 503
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
here, a = 25, b = -75, c = 50
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (25 – 75 + 50)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = 3abc
253 + (-75)3 + 503 = 3abc
= 3(25)(-75)(50)
= -281250
Hence, the value 253 + (-75)3 + 503 = -281250
(b) 483 – 303 – 183
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
here, a = 48, b = -30, c = -18
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (48 – 30 – 18)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = 3abc
483 + (-30)3 + (-18)3 = 3abc
= 3(48)(-30)(-18)
= 77760
Hence, the value 483 + (-30)3 + (-18)3 = 77760
(c) \((\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}\)
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
here, a= \(\frac{1}{2}\)
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (\(\frac{1}{2}\)
by using least common multiple
a3 + b3 + c3 = \((\frac{1*6}{2*6} + \frac{1*4}{3*4} – \frac{5*2}{6*2})\)
a3 + b3 + c3 = \((\frac{6}{12} + \frac{4}{12} – \frac{10}{12})\)
a3 + b3 + c3 = 0( a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = 3abc
(\(\frac{1}{2}\)
= \(\frac{1}{2}\)
= \(\frac{-5}{12}\)
Hence, the value of \((\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}\)
(d) (0.2)3 – (0.3)3 + (0.1)3
we know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
here, a = 0.2, b = 0.3, c = 0.1
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (0.2 – 0.3 + 0.1)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = 3abc
(0.2)3 – (0.3)3 + (0.1)3 = 3abc
= 3(0.2)(-0.3)(0.1)
= -0.018
Hence, the value (0.2)3 – (0.3)3 + (0.1)3 is 0.018