Question 147

If $$a^3+b^3=416$$ and $$a+b=16$$, then $$(a-b)^{2}+ab$$ is equal to:

Solution

$$(a+b)^3-3ab(a+b)=a^3+b^3$$

So, $$(a+b)^3-3ab(a+b)=416$$

or,$$16^3-3\times ab\times 16=416$$

or,$$3680=3\times ab\times 16$$

or,$$3ab=230.$$

or,$$ab=230/3.$$

And

$$(a-b)^2=(a+b)^2-4ab.$$

or,$$(a-b)^2=16^2-920/3$$

or,$$(a-b)^2=-152/3.$$

So, $$(a-b)^2+ab=(230-152)/3=78/3=26.$$

So, D is correct.

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