Question

x+ 1/x = 3 ; x > 1


Prove that,
x⁵ + 1/(x⁵) = 123​

Answer 1

Answer:

Assuming the question as (x + 1/x) = 3

Squaring on both the sides we get

(x + 1/x)2 = 32

x2 +(1/x2) + 2 = 9

⇒ x2 +(1/x2) = 7

Now cubing on both the sides we get,

[x2 +(1/x2)]3 = 73

LHs is in the form of (a + b)3 = a3 + b3 + 3ab (a + b)

Hence [x2]3 + [(1/x2)]3 + 3 (x2) × (1/x2)[x2 +(1/x2)] = 343

⇒ x6 + (1/x6) + 3 × 7 = 343

⇒ x6 + (1/x6) + 21 = 343

∴ x6 + (1/x6) = 343 − 21 = 322

Answer 2

We know the polynomial identity,

$\hookrightarrow (a+b)^{2}=a^{2}+2ab+b^{2}$

So,

$\hookrightarrow\left(x+\dfrac{1}{x}\right)^{2}=x^{2}+2+\dfrac{1}{x^{2}}$

$\hookrightarrow3^{2}-2=x^{2}+\dfrac{1}{x^{2}}$

$\hookrightarrow\red{\boxed{\red{x^{2}+\dfrac{1}{x^{2}}=7}}}$

Also, we know the polynomial identity,

$\hookrightarrow(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

So,

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

$\hookrightarrow3^{3}-9=x^{3}+\dfrac{1}{x^{3}}$

$\hookrightarrow\red{\boxed{\red{x^{3}+\dfrac{1}{x^{3}}=18}}}$

From above,

$\hookrightarrow\left(x^{2}+\dfrac{1}{x^{2}}\right)\left(x^{3}+\dfrac{1}{x^{3}}\right)=126$

$\hookrightarrow x^{5}+\dfrac{1}{x}+x+\dfrac{1}{x^{5}}=126$

$\hookrightarrow x^{5}+3+\dfrac{1}{x^{5}}=126$

$\hookrightarrow\red{\boxed{\red{x^{5}+\dfrac{1}{x^{5}}=\red{\underline{123}}}}}$

Hence, it is now proven that the value of $x^{5}+\dfrac{1}{x^{5}}$ is 123.