Question

Find the value of 64 x cube + 27 y cube if 4 x + 3 y is equal to 5 and x y is equal to 1

Answer 1

Answer:answer would be -55.

Now, (4x+3y)^2=(4x)^2+(3y)^2+2*4x*3y

(5)^2=(4x)^2+(3y)^2+2*4*3*xy

25=(4x)^2+(3y)^2+24*1

25=(4x)^2+(3y)^2+24

(4x)^2+(3y)^2=25-24

(4x)^2+(3y)^2=1. ........(1)

Now, 64x^3+27y^3=(4x)^3+(3y)^3

(4x)^3+(3y)^3=(4x+3y){(4x)^2+(3y)^2-4x*3y}

Since , a^3+b^3=(a+b)(a^2+b^2-ab}

(4x)^3+(3y)^3=(5){1-4x*3y}

(4x)^3+(3y)^3=(5)(1-4*3*xy)

(4x)^3+(3y)^3=(5)(1-12)

(4x)^3+(3y)^3=5*-11

(4x)^3+(3y)^3=-55

64x^3+27y^3=-55.

Thus , ans is -55 .....