Q8 of 59 Page 7

If x – y = – 6 and xy = 4, find the value of x3 – y3

The identity used here is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ (x – y)3 = x3 + 3x2(– y) + 3x(– y)2 + (– y)3


⇒ x3 – 3x2y + 3xy2 – y3


⇒ (x – y)3 = x3 – 3x2y + 3xy2 – y3


⇒ (– 6)3 = x3 – y3 + 3xy(– x + y)


⇒ – 216 + 3 × 4 (– 6) = x3 – y3


⇒ – 216 – 72 = x3 – y3


⇒ – 288 = x3 – y3


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