Question 66

If a + b + c = 7 and ab + bc + ca = -6, then the value of $$a^3 + b^3 + c^3 - 3abc$$ is:

Solution

We know that,

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

a + b + c = 7

Squaring both sides,

$$(a + b + c)^2 = 49$$

$$a^2 + b^2 + c^2 + 2(ab + bc + ac) = 49$$

$$a^2 + b^2 + c^2 = 49 + 12 = 61$$

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

= 7(61 - (-6)) = 7 $$\times$$ 67 = 469

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