Factorise p3+27q3+8r3-18pqr
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Let p(q - r) = a, q(r - p) = b and r(p - q) = c,
=> a + b + c = 0
=> a + b = - c
=> (a + b)^3 = - c^3
=> a^3 + b^3 + 3ab(a + b) + c^3 = 0
=> a^3 + b^3 + c^3 = 3abc (because a + b = - c)
Now, p3(q – r)3 + q3(r – p)3 + r3(p – q)3 =
a^3 + b^3 + c^
= 3abc
= 3pqr(p - q)(q - r)(r - p)
=> a + b + c = 0
=> a + b = - c
=> (a + b)^3 = - c^3
=> a^3 + b^3 + 3ab(a + b) + c^3 = 0
=> a^3 + b^3 + c^3 = 3abc (because a + b = - c)
Now, p3(q – r)3 + q3(r – p)3 + r3(p – q)3 =
a^3 + b^3 + c^
= 3abc
= 3pqr(p - q)(q - r)(r - p)
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