Question

if 4a+3b =10 and ab=2 ,find the value of 64a^3+27b^3

Answer 1

hey friend ,

here we have to find the value of 64a^3 +27b^3 so we can see that 64 is the cube of 4 and 27 is cube of 3.

Given :-

=) (4a + 3b) =10 --------(1)

cubing both sides.

=) (4a+3b)^3 = (10)^3

=)64a^3 +27b^3 + 3×4a×3b(4a+3b) =1000

=)64a^3 +27b^3 +36ab(4a+3b) =1000---(2)

so also given that :---

=) ab = 2. and 4a +3b = 10

put these value in equation (2) we get.

=) 64a^3 +27b^3 +36×2×10 =1000

=) 64a^3 +27b^3 + 720 =1000

=) 64a^3 + 27b^3 = 1000 - 720

=) 64a^3 +27b^3 = 280

so the required value is 280 ans.
______________________________

hope it will help u

Answer 2

given,
4a+3b=10
squaring on both sides,
(4a+3b)²=(10)²
⇒(4a)²+(3b)²+2(4a)(3b)=100
⇒(4a)²+(3b)²=100-24ab
⇒(4a)²+(3b)²=100-24×2
⇒(4a)²+(3b)²=100-48=52     ---------------  1

64a³+27b³=(4a)³+(3b)³
[ x³+y³=(x+y)(x²+xy+y²) ]
⇒ (4a)³+(3b)³ = (4a+3b)[(4a)²+(4a×3b)+(3b)²]
                      = 10(52+12ab)     [from 1]
                      = 10(52+12×2)
                      = 10×76
                      = 760