There are three brothers. Two of them are twins.The age of one brother is cube of the other. Next year the age of one brother would be square of the other. What is the sum of their current ages?
+128407 Let the twins' current ages = A
Let the current age of the other brother = A^3
And next year we have that
A^3 + 1 = ( A + 1)^2
A^3 + 1 = A^2 + 2A + 1
A^3 - A^2 - 2A = 0
A(A^2 - A - 2) = 0
One solution is that A =0 (reject)
And using the second factor we have that
A^2 - A - 2 = 0
(A - 2) ( A + 1) = 0
Setting both factors to 0 and solving for A we have that
A = 2 or A = -1 (reject)
So
A= 2 and the non-twin = 2^3 = 8
So.....the sum of the ages =
2 + 2 + 8 =
12
