Question

Prove the following identity:
${(b+c)}^3+{(c+a)}^3+{(a+b)}^3-3(b+c)(c+a)(a+b)=2(a^3+b^3+c^3-3abc)$.

Answer 1

L.H.S$=(b+c{)}^3+(c+a{)}^3+(a+b{)}^3-3(b+c\left)\right(c+a\left)\right(a+b)$

$=b^3+c^3+3b^{2}c+3b{c}^2+c^3+a^3+3c^{2}a+3c{a}^2+a^3+b^3+3a^{2}b+3a{b}^2-3(b+c)(c+a)(a+b)$

$=2(a^3+b^3+c^3)+3b^{2}c+3b{c}^2+3c^{2}a+3c{a}^2+3a^{2}b+3a{b}^2-3(b+c)(c+a)(a+b)+6abc-6abc$

$=2(a^3+b^3+c^3)+3(b^{2}c+b{c}^2+c^{2}a+c{a}^2+a^{2}b+a{b}^2+2abc)-3(b+c)(c+a)(a+b)-6abc$

$=2(a^3+b^3+c^3)+3(b+c)(c+a)(a+b)-3(b+c)(c+a)(a+b)-6abc$

$=2(a^3+b^3+c^3)-6abc$

$=2(a^3+b^3+c^3-3abc)$

$=$R.H.S