Question

If x^2+1/x^2=34,find x^3+1/x^3-9

Answer 1

189

Step-by-step explanation:

${\underline{{\text{Given:}}}}$

${x}^{2} + \dfrac{1}{ {x}^{2} } = 34$

${\underline{{\text{To Find:}}}}$

${x}^{3} + \dfrac{1}{ {x}^{3} } - 9$

${\underline{{\text{Solution:}}}}$

First we should find the value of $x+ \dfrac{1}{x}$

So,

${x}^{2} + \frac{1 }{ {x}^{2} } = 34 \\ \\ [\text{Adding 2 in both side}] \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 34 + 2 \\ \\ {x}^{2} + 2. x. \frac{1}{x} + \frac{ 1 }{ {x}^{2} } = 36 \:\: [\text{Using i}]\\ \\ {(x + \frac{1}{x}) }^{2} = {(6)}^{2} \\ \\ [\text{Cancelling power in both side}] \\ \\ x + \frac{1}{x} = 6 \: \: \: \: .......(1)$

Now,

$x + \frac{1}{x} = 6 \\ \\ [\text{Cubing both side}] \\ \\ {(x + \frac{1}{x})}^{3} = {6}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.x. \frac{1}{x} (x + \frac{1}{x} ) = 216 \: \: [\text{Using ii}] \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3. \cancel{x}. \frac{1}{\cancel{x}} (6) = 216 \: [\text{From 1}] \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 18 = 216 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 216 - 18 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 198 \\ \\ [\text{Subtracting 9 from both side}] \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } - 9= 198 - 9 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } - 9= 189$

Hence, the required answer is 189.

${\boxed{\blue{\text{Important Algebra Formula}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy....[\text{i}]\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y)....[\text{ii}]\\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$