Question

If (4x + 3y) = 8 and xy = 1, find the value of (64x3 + 27y3).
urgent

Answer 1

Answer:

Answer:

64x^3+27y^3

Can be represented as :

(4x)^3+(3y)^3

by using identity:

(a+b) (a^2-ab+b^2):

(4x+3y) (4x^2-4x×3y+3y^2)

(8) [4x^2-12(1)+3y^2]... .. ... (1) ( given :(4x+3y)=8 and xy =1)

b

Now finding the value of (4x^2+3y^2)

Using Identity (a^2+b^2)= (a+b)^2-2ab:

(4x^2+3y^2)= (4x+3y )^2 -2×4x×3y

=( 8)^2 -24xy

= (64) -24(1)

= 40...........(2)

Putting the value of (2) in (1):

=(8) [4x^2+3y^2-12xy]

=(8) [ 40 -12(1)]

=(8) [ 28]

=224