Question 69

If a + b + c = 6 and ab + bc + ca = 4, then $$a^3 + b^3 + c^3 - 3 abc$$ is equal to:

Solution

$$\left(a+b+c\right)=6$$

Squaring both side,

$$\left(a+b+c\right)^2=6^2$$

$$a^2+b^2+c^2+2\left(ab+bc+ac\right)=36$$

$$a^2+b^2+c^2+2\left(4\right)=36$$

$$a^2+b^2+c^2=28$$

Now,

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

= $$6\left(28-4\right)=144$$

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