Bunuel wrote:
If a and b are integers and \((\sqrt[3]{a}*\sqrt{b})^6 = 500\), then a + b could equal
A. 2
B. 3
C. 4
D. 5
E. 6
From the above equation we get,
\([(a^{1/3}) X (b^{1/2})]^6 = 500\)
or\(a^2 X b^3 = 5^3 X 2^2\)
Thus we can write as , a= 2 and b = 5.
But a+b = 7 that is not in the option. However a could be = 2 or -2 (since square of any negative number is positive)
that is \(2^2 = (-2)^2\)
Now when a= -2 and b= 5, we have = -2 + 5 =3 . Hence option B