Question
Find: $\mathrm{a}^3-\frac{1}{\mathrm{a}^3}$, if $\mathrm{a}-\frac{1}{\mathrm{a}}=4$.
✓
Answer
$ a-\frac{1}{a}=4$
$ \Rightarrow\left(a-\frac{1}{a}\right)^3=(4)^3$
$ \Rightarrow a^3-\frac{1}{a^3}-3 a \times \frac{1}{a}\left(a-\frac{1}{a}\right)=64$
$ \Rightarrow a^3-\frac{1}{a^3}+3(4)=64 \ldots \ldots . . .\left[\because-\frac{1}{a}=4\right]$
$ \Rightarrow a^3-\frac{1}{a^3}-12=64$
$ \Rightarrow a^3-\frac{1}{a^3}=64+12$
$ \Rightarrow a^3-\frac{1}{a^3}=76$
$ \Rightarrow\left(a-\frac{1}{a}\right)^3=(4)^3$
$ \Rightarrow a^3-\frac{1}{a^3}-3 a \times \frac{1}{a}\left(a-\frac{1}{a}\right)=64$
$ \Rightarrow a^3-\frac{1}{a^3}+3(4)=64 \ldots \ldots . . .\left[\because-\frac{1}{a}=4\right]$
$ \Rightarrow a^3-\frac{1}{a^3}-12=64$
$ \Rightarrow a^3-\frac{1}{a^3}=64+12$
$ \Rightarrow a^3-\frac{1}{a^3}=76$
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