Question

If u + 1/u = 8, find the value of : -

(i) u2

+ 1/u2

(ii) u3

+ 1/u3​

Answer 1

Answer:

The value of $u^2+\frac{1}{u^2}$ is 62 and the value of $u^3+\frac{1}{u^3}$ is 488.

Step-by-step explanation:

Given that

$u+\frac{1}{u}=8$

i) We can first find the expansion of the following expression.

$(u+\frac{1}{u})^2=u^2+2\times u\times \frac{1}{u}+\frac{1}{u^2} =u^2+2+\frac{1}{u^2}$

Thus

$\implies u^2+\frac{1}{u^2} = (u+\frac{1}{u})^2-2$

We can substitute the value of $u+\frac{1}{u}$,

$\implies u^2+\frac{1}{u^2} =8^2-2=64-2=62$

ii) We have the expansion,

$(u+\frac{1}{u})^3=u^3+3\times u^2\times \frac{1}{u}+3\times u \times \frac{1}{u^2} + \frac{1}{u^3} =u^3+3u+\frac{3}{u}+ \frac{1}{u^3}\\\\ (u+\frac{1}{u})^3= u^3+ \frac{1}{u^3}+3(u+\frac{1}{u})$

So we have

$\implies u^3+\frac{1}{u^3}=(u+\frac{1}{u})^3-3(u+\frac{1}{u} )$

On substituting the values, we get

$u^3+\frac{1}{u^3}=8^3-3\times 8=488$