Question

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

Answer 1

Answer:

64x^3 +27y^3 is 224

Step-by-step explanation:

given 4x +3y = 8 , xy = 1

asked to find the value of 64x^3 + 27y^3

4x + 3y = 8 ----------(1) , xy = 1 ---------(2)

64 can be written as 4 cube and 27 can be written as 3

cube so we can cube equation (1) on both sides

(a+b)^3 = a^3+b^3 + 3ab(a+b)

(4x +3y)^3 = 8^3

64x^3 + 27y^3 +3(4x)(3y)(4x+3y) = 512

64x^3 + 27y^3 + 36(xy) ( 8) = 512 ( from equation (1) 4x+3y

is 8)

64x^3+27y^3+ 36(1)(8) = 512( from equation (2) xy =1)

64x^3+27y^3 + 288 = 512

64x^3 + 27y^3 = 512 - 288

64x^3 + 27y^3 = 224

Answer 2

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