Question

3√((x+1)(y+2)(z+3))= 7
Find xyz.

Answer 1

3√(x+1)(y+2)(z+3)=7
so
(x+1)(y+2)(z+3)=7×7×7
for this case
x+1=7
y+2=7
z+3=7
because 7 is a prime number and it don't have any other multiples or factors
by this we can find x,y and z
and multiply them

Answer 2

³√(x+1)(y+2)(z+3)= 7
(x+1)(y+2)(z+3) = 7³

(x+1) (y+2)(z+3) = 7×7×7

=> x+1 = 7
=> x = 6

y+2 = 7
=> y = 5

z+3 = 7
=> z= 4

so , xyz = (6)(5)(4) => 30×4 = 120


hope this helps