# If x4 + 1/x4 = 194, calculate

Question:

If $x^{4}+1 / x^{4}=194$, calculate $x^{2}+1 / x^{2}, x^{3}+1 / x^{3}, x+1 / x$

Solution:

Given,

$x^{4}+1 / x^{4}=194$

add and subtract $\left(2^{*} x^{2} * 1 / x^{2}\right)$ on left side in above given equation

$x^{4}+1 / x^{4}+\left(2^{*} x^{2} * 1 / x^{2}\right)-2\left(2^{*} x^{2} * 1 / x^{2}\right)=194$

$x^{4}+1 / x^{4}+\left(2^{*} x^{2} * 1 / x^{2}\right)-2=194$

$\left(x^{2}+1 / x^{2}\right)^{2}-2=194$

$\left(x^{2}+1 / x^{2}\right)^{2}=194+2$

$\left(x^{2}+1 / x^{2}\right)^{2}=196$

$\left(x^{2}+\frac{1}{x^{2}}\right)=\sqrt{196}$

$\left(x^{2}+1 / x^{2}\right)=14 \ldots 2$

Add and subtract (2*x* 1/x) on left side in eq 2

$\left(x^{2}+1 / x^{2}\right)+\left(2^{*} x^{*} 1 / x\right)-\left(2^{*} x^{*} 1 / x\right)=14$

$(x+1 / x)^{2}-2=14$

$(x+1 / x)^{2}=14+2$

$(x+1 / x)^{2}=16$

$(x+1 / x)=\sqrt{16}$

$(x+1 / x)=4 \ldots 3$

Now, cubing eq 3 on both sides

$(x+1 / x)^{3}=4^{3}$

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

$x^{3}+1 / x^{3}+3^{*} x^{*} 1 / x(x+1 / x)=64$

$x^{3}+1 / x^{3}+\left(3^{*} 4\right)=64$

$x^{3}+1 / x^{3}=64-12$

$x^{3}+1 / x^{3}=52$

hence, the values of $\left(x^{2}+1 / x^{2}\right)^{2}=196,(x+1 / x)=4, x^{3}+1 / x^{3}=52$