Question 106

If $$a + b + c = 19, ab + bc + ca = 120$$, then what is the value of $$a^3 + b^3 + c^3 - 3abc$$?

Solution

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

$$ 19^2 = a^2 +b^2 + c^2 + 2(120)$$

$$a^2 +b^2 + c^2 = 361 - 240 = 121$$

$$a^3 + b^3 + c^3 - 3abc$$

=$$ (a + b + c)(a^2 + b^2 + c^2 -ab -bc - ca)$$

= 19(121$$ \times -120) = 19$$

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