Question

If $(m^2+\frac{1}{m^2})=51$, find the value of $(m^3-\frac{1}{m^3})$.

Answer 1

Step 1: Find the value of $(m-\frac{1}{m})$
Given the value of expression$(m^2+\frac{1}{m^2})=51\cdots \left(1\right)$
Let's assume$(m-\frac{1}{m})=p$.
Now, find the square on both the side by using the formula$(a-b{)}^2=a^2+b^2-2ab$
$\Rightarrow {(m-\frac{1}{m})}^2=p^2$
$\Rightarrow m^2+\frac{1}{m^2}-2\left(m\right)\left(\frac{1}{m}\right)=p^2$
$\Rightarrow 51-2=p^2$
$\Rightarrow p^2=49$
$\therefore p=(m-\frac{1}{m})=\pm 7$ ...(2)

Step 2: Find the Value of $(m^3-\frac{1}{m^3})$
Now, find the cube of equation (2) by using the formula
$(a-b{)}^3=a^3-b^3-3ab(a-b)$
$\Rightarrow {(m-\frac{1}{m})}^3=(m{)}^3-{\left(\frac{1}{m}\right)}^3-3(m)\left(\frac{1}{m}\right)(m-\frac{1}{m})$
$\Rightarrow (\pm 7{)}^3=m^3-\frac{1}{m^3}-3(\pm 7)$
$\therefore m^3-\frac{1}{m^3}=\pm (343+21)=\pm 364$
Hence, the value of $(m^3-\frac{1}{m^3})$ is $\pm 364$.