Question

(p $-$ 3q)³ + (3q $-$ 7r)³ + (7r $-$ 9)³

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}{\left(p-3q\right)}^3+{\left(3q-7r\right)}^3+{\left(7r-p\right)}^3 \\ \mathrm{Let}p-3q=a,3q-7r=b\mathrm{and}7r-p=c \\ \therefore {\left(p-3q\right)}^3+{\left(3q-7r\right)}^3+{\left(7r-p\right)}^3=a^3+b^3+c^3 \\ \mathrm{Here},a+b+c=p-3q+3q-7r+7r-p=0 \\ \mathrm{However},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{if}a+b+c=0,\mathrm{then}{a}^3+b^3+c^3=3abc \\ \therefore {\left(p-3q\right)}^3+{\left(3q-7r\right)}^3+{\left(7r-p\right)}^3=3\left(p-3q\right)\left(3q-7r\right)\left(7r-p\right) \end{gathered}$$