factorise: p√p + 8q√q + 27r√r -18√pqr
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= p√p + 8 q√q + 27 r√r - 18 √( pqr )
= ( √p )^3 + ( 2√q )^3 + ( 3√r )^3 - 3 * √p * √q * √r
By using identity :
a^3 + b^3 + c^3 - 3 abc = ( a + b + c ) ( a^2 + b^2 + c^2 - ab - bc - ca )
=( √p + √q + √r ) { (√p)^2 + (2√q)^2 + ( 3√r )^2 - √p*2√q - 2√q*3√r - 3√r*√p }
= ( √p + √q + √r ) ( p + 4 q + 3 r - 2√pq - 6√qr - 3 √pr )
= ( √p )^3 + ( 2√q )^3 + ( 3√r )^3 - 3 * √p * √q * √r
By using identity :
a^3 + b^3 + c^3 - 3 abc = ( a + b + c ) ( a^2 + b^2 + c^2 - ab - bc - ca )
=( √p + √q + √r ) { (√p)^2 + (2√q)^2 + ( 3√r )^2 - √p*2√q - 2√q*3√r - 3√r*√p }
= ( √p + √q + √r ) ( p + 4 q + 3 r - 2√pq - 6√qr - 3 √pr )
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