Question

I hope now it is clear now tell me question 25​

Answer 1

Answer:

6

Step-by-step explanation:

thank me if u get it

and wait

Answer 2

Taking a =x

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

$\tt{X+\dfrac{1}{X}}$=√8 + 3

${\sf{\green{\underline{\large{To\:find}}}}}$

$\tt{X^3+\dfrac{1}{X^3}}$

$\tt{X^2+\dfrac{1}{X^2}}$

$\tt{X^3+\dfrac{1}{X^3}}$

$\tt{X^4+\dfrac{1}{X^4}}$

${\sf{\pink{\underline{\Large{Explanation}}}}}$

x= 3 +√8

Now 1/x =

$\tt\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}$

$\rightarrow\tt\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}$

=> 1/x = 3-√8

Now

x + 1/x = 3+√8 + 3-√8

=>x + 1/x = 6

$\boxed{\boxed{\sf\red{(a+b)^2=a^2+b^2+2ab}}}$

Therefore

$\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

Solving

$\tt{X+\dfrac{1}{X}}$=6

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(6)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=36$

$\implies\tt{36=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{36=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{36-2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{34=X^2+\dfrac{1}{X^2}}$

Now

$\tt{X^3+\dfrac{1}{X^3}}$

Formula used

$\implies\tt{X^3+\dfrac{1}{X^3)}=(X+\dfrac{1}{x})(X^2+\dfrac{1}{X^2}-\frac{1}{X}\times{X)}}$

¶utting value

$\implies\tt{X^3+\dfrac{1}{X^3}={6}(34-1)}$

$\implies\tt{X^3+\dfrac{1}{X^3}=6\times{33}}$

$\implies\tt{X^3+\dfrac{1}{X^3}=198}$

Again

$\implies\tt{34=X^2+\dfrac{1}{X^2}}$

Both side squaring

=>x⁴ + 1/ x⁴ = 1156 -2

=>x⁴ + 1/ x⁴ = 1154