Question

if x^2+1/x^2=7 then find the value of x^3+1/x^3

Answer 1

(x + 1/x) = x + 1/x + 2
 (x + 1/x) = 9
x + 1/x = 3 
now
 (x + 1/x) = x + 1/x + 3(x + 1/x)
27 = x + 1/x + 9
or
 x + 1/x = 27 - 9 = 18
it should be the answer 

Answer 2

The value of $x^{3}+\frac{1}{x^{3}} \text { is } \bold{18 \text { or }-18}$

Solution:

Given $x^{2}+\frac{1}{x^{2}}=7$

We know that $(a+b)^{2}=a^{2}+2 a b+b^{2}$

Similarly, $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2 \times x \times \frac{1}{x}$

$=x^{2}+\frac{1}{x^{2}}+2$

$=7+2=9$

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=9$

$\Rightarrow x+\frac{1}{x}=3 \text { or }-3$     

Case I : $\left(x+\frac{1}{x}\right)=3$  

We know that $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

Similarly, $\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3 \times x^{2} \times \frac{1}{x}+3 \times x \times \frac{1}{x^{2}}$

$=x^{3}+\frac{1}{x^{3}}+3 x+\frac{3}{x}$

$=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)$

$x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)$

$=3 \times 3 \times 3-3 \times 3$

$=27-9=18$

Case II : $\left(x+\frac{1}{x}\right)=-3$

$x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)$

$=(-3) \times(-3) \times(-3)-3 \times(-3)$

$=-27+9=-18$