Question 85

If $$x + y + z = 6, xyz = -10$$ and $$x^2 + y^2 + z^2 = 30$$, then what is the value of $$(x^3 + y^3 + z^3)$$?

Solution

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(ab + bc + ac)$$

$$6^2 = 30 + 2(ab + bc + ac)$$

(ab + bc + ac) = 3

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

$$x^3 + y^3 + z^3 = 6 \times (30 - 3) + 3 \times (-10)$$

$$x^3 + y^3 + z^3 = 162 - 30 = 132$$

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