Question

find the value :
27a^3+8b^3 if 3a+2b=6 and ab=2

Answer 1

if any problem comment on it

Answer 2

Given are the values of '3a+2b' and 'ab', find the value of $27a^3+8b^2$.

Explanation:

• Let there be two real numbers 'x' and 'y'.
• Then using the identity for the cube of the sum of two numbers we have, $(x+y)^3=x^3+3a^2b+3ab^2+y^3$  
• Now here let's put $x=3a,\ y=2b$  
• Then we get,
•            $->(3a+2b)^3=27a^3+3(3a)^2(2b)+3(3a)(2b)^2+8b^3 \\->(3a+2b)^3=27a^3+8b^3+54a^2b+36ab^2\\->(3a+2b)^3=27a^3+8b^3+18ab(3a+2b) \\->27a^3+8b^3=(3a+2b)^3-18ab(3a+2b)$ ---(a)
• Now it is given that $3a+2b=6, \ \ ab=2$  
• Putting these values in (a) we get,
•           $->27a^3+8b^3=(6)^3-18(2)(6)\\->27a^3+8b^3=216-216\\->27a^3+8b^3=0$  
• Hence the value the required equation is found to be $0$.