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If a + b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc. |
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Answer» a + b + c = 9, ab + bc + ca = 26 Squaring, a + b + c = 9 both sides, we get (a + b + c)2 = (9)2 a2 + b2 + c2 + 2(ab + bc + ca) = 81 a2 + b2 + c2 + 2 x 26 = 81 a2 + b2 + c2 + 52 = 81 a2 + b2 + c2 = 29 Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)] = 9[29 – 26] = 9 x 3 = 27 ⇒ a3 + b3 + c3 – 3abc = 27 |
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