If p³ + q³ + r³ - 3pqr = 4, and a = q + r, b = r + p and c = p + q, then what is the value of a³ + b³ + c³ - 3abc?

Solution (By Examveda Team)

$$\eqalign{ & {p^3} + {q^3} + {r^3} - 3pqr = 4 \cr & a = q + r \cr & b = r + p \cr & c = p + q \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc \cr & = {\left( {q + r} \right)^3} + {\left( {r + p} \right)^3} + {\left( {p + q} \right)^3} - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = {q^3} + {r^3} + 3qr\left( {q + r} \right) + {r^3} + {p^3} + 3rp\left( {r + p} \right) + {p^3} + {q^3} + 3pq\left( {p + q} \right) - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = 2{p^3} + 2{q^3} + 2{r^3} + 3pq\left( {p + q} \right) + 3qr\left( {q + r} \right) + 3rp\left( {r + p} \right) - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = 2\left( {{p^3} + {q^3} + {r^3}} \right) + 3{p^2}q + 3p{q^2} + 3{p^2}r + 3q{r^2} + 3{r^2}p + 3r{p^2} - 3\left[ {\left( {p + q} \right)\left( {qr + qp + {r^2} + rp} \right)} \right] \cr & = 2\left( {4 + 3pqr} \right) + 3{p^2}q + 3p{q^2} + 3{q^2}r + 3q{r^2} + 3{r^2}p + 3r{p^2} - 3pqr - 3{p^2}q - 3p{r^2} - 3{p^2}r - 3{q^2}r - 3q{r^2} - 3{q^2}p - 3pqr \cr & = 8 + 6pqr - 6pqr \cr & = 8 \cr & \cr & {\bf{Alternate:}} \cr & r = q = 0 \cr & {p^3} + 0 + 0 - 0 = 4 \cr & {p^3} = 4,\,a = 0,\,b = p,\,c = p \cr & {a^3} + {b^3} + {c^3} - 3abc \cr & = 0 + {p^3} + {p^3} - 0 \cr & = 2{p^3} \cr & = 2 \times 4 \cr & = 8 \cr} $$