Question 74

If $$a^2 + b^2 + c^2 + 27 = 6( a + b + c)$$, then what is the value of $$\sqrt[3]{a^3 + b^3 - c^3}$$

Solution

$$a^2 + b^2 + c^2 + 27 = 6( a + b + c)$$

$$(a^2 -6a+9) + (b^2 -6b+9) +(c^2 -6c+9)=0$$

$$(a-3)^2+(b-3)^2+(c-3)^2=0 $$which, in turn, implies that a=3, b=3 and c=3

$$\left(a^3+b^3-c^3\right)^{\frac{1}{3}}=27+27-27=27$$

$$\sqrt[3]{a^3 + b^3 - c^3}=\sqrt[\ 3]{27}=3$$

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