Question

f (x-1/8)=8,find the value of (x²+1/x²) and (x⁴+1/x⁴)​

Answer 1

Your question is wrong.

It's $\sf x - \dfrac{1}{x}$ instead of $\sf x - \dfrac{1}{8}$

Value of x can be easily founded in your question so your question is wrong.

★ɢɪᴠᴇɴ:-

$\sf x - \dfrac{1}{x} = 8$

★ᴛᴏ ғɪɴᴅ:-

• $\sf x^2 + \dfrac{1}{x^2}$

• $\sf x^4 + \dfrac{1}{x^4}$

★sᴏʟᴜᴛɪᴏɴ:-

We know that,

(a-b)² = a²+b²-2ab

Using this identity,

$\implies \sf \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}$

$\implies \sf (8 )^2 = x^2 + \dfrac{1}{x^2} - 2$

$\implies \sf 64= x^2 + \dfrac{1}{x^2}-2$

$\implies \sf 64+2= x^2 + \dfrac{1}{x^2}$

$\implies \sf 66 = x^2 + \dfrac{1}{x^2}$

______________________________

Now we will find $\sf x^4 + \dfrac{1}{x^4}$

We know that,

(a+b)² = a²+b²+2ab

Using this identity,

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

$\implies \sf (66)^2 = x^4 + \dfrac{1}{x^4} + 2$

$\implies \sf 4356 = x^4 + \dfrac{1}{x^4} + 2$

$\implies \sf 4356 - 2= x^4 + \dfrac{1}{x^4}$

$\implies \sf 4354= x^4 + \dfrac{1}{x^4}$

Hope it helps