Question
If $2 x-\frac{1}{2 x}=4$, find : $8 x^3-\frac{1}{8 x^3}$
✓
Answer
$ 2 \mathrm{x}-\frac{1}{2 \mathrm{x}}=4$
$ \Rightarrow\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^3=(4)^3$
$ \Rightarrow(2 \mathrm{x})^3-\left(\frac{1}{2 \mathrm{x}}\right)^3-3 \times 2 \mathrm{x} \times \frac{1}{2 \mathrm{x}}\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-3(4)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-12=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=64+12$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=76$
$ \Rightarrow\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^3=(4)^3$
$ \Rightarrow(2 \mathrm{x})^3-\left(\frac{1}{2 \mathrm{x}}\right)^3-3 \times 2 \mathrm{x} \times \frac{1}{2 \mathrm{x}}\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-3(4)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-12=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=64+12$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=76$
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