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
_________________
If you found this post useful, please let me know by pressing the Kudos ButtonRules for PostingGot 20 Kudos? You can get Free GRE Prep Club TestsGRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months
GRE Prep Club tests