Question 55
$$a + b + c = 6, a^{2} + b^{2} + c^{2} = 14$$ and $$a^{3} + b^{3} + c^{3} = 36$$, then the value of abc is:
Solution
a + b + c = 6
Squaring on both sides:
$$a^2+b^2+c^2+2\left(ab\ +\ bc\ +\ ca\right)=36$$.
$$a^{2} + b^{2} + c^{2} = 14$$
This gives: ab + bc + ca = 11.
$$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$$
Putting in all the given values, we get:
36 - 3abc = 6(14 - 11)
abc = 6.
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