Question

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

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{x^{2}+\dfrac{1}{x^{2}}=7}\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto \sf{x^{3}+\dfrac{1}{x^{3}}}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

We have,

$\sf{\red{\mapsto x^{2}+\dfrac{1}{x^{2}}=7}}$

$\sf{\implies x^{2}+\dfrac{1}{x^{2}}=7}$

$\sf{\implies x^{2}+\dfrac{1}{x^{2}}+2\times x\times \dfrac{1}{x}=7+2\times x\times \dfrac{1}{x}}$

$\sf{\implies (x+\dfrac{1}{x}) ^{2}=7+2}$

$\large\purple{\boxed{\bf{\orange{(a+b) ^{2}=a^{2}+b^{2}+2ab}}}}$

$\sf{\implies (x+\dfrac{1}{x}) ^{2}=9}$

$\sf{\implies (x+\dfrac{1}{x}) ^{2}=(3)^{2}}$

$\sf{\implies x+\dfrac{1}{x}=3}$

${\underline{\red{\sf{(\leadsto Ignoring\:negative\:value\:of\:3)}}}}$

______________________________________

$\purple{\bf{\mapsto Cubing\:both\:sides}}$

$\sf{\implies (x+\dfrac{1}{x}) ^{3}=3^{3}}$

$\sf{\implies x^{3}+\dfrac{1}{x^{3}}+\dfrac{3x}{x}(x+\dfrac{1}{x}) =27}$

$\sf{\implies x^{3}+\dfrac{1}{x^{3}}+3\times 3=27}$

$\sf{\implies x^{3}+\dfrac{1}{x^{3}} =27-9}$

${\underline{\pink {\bf{\mapsto x^{3}+\dfrac{1}{x^{3}}=18}}}}$

Answer 2

Answer:

Given Equation is x^2 + 1/x^2 = 7

We know that (x + 1/x)^2 = x^2 + 1/x^2 + 2 * x * 1/x

                                         = x^2 + 1/x^2 + 2

                                        = 7 + 2

                                        = 9

                          (x + 1/x) = 3

Now,

(x^3 + 1/x^3) = (x + 1/x)^3 - 3 * x * 1/x * (x + 1/x)

                     = (x + 1/x)^3 - 3(x + 1/x)

 

                    = (3)^3 - 3(3)

                   = 27 - 9

                   = 18.

Hope this helps!