Question

Find the value of 64x3 +343y3, if 4x + 7y=12 and xy=1/21.​

Answer 1

Answer:

1680

Step-by-step explanation:

given

4x + 7y = 12

xy = 1/21

now,

64x3 + 343y3

= (4x)3 + (7y)3

= (4x + 7y)3 - 3*4x*7y(4x + 7y)

= (12)3 - 84xy*12

= (12)3 - 84*12/21

= 1728 - 48

= 1680

Answer 2

Step-by-step explanation:

64x^3+343y^3

we can write this Equation as:

(4x)^3+(7y)^3

Now according to Identity

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

Putting the values

(4x+7y)^3=(4x)^3+(7y)^3+3×4x×7y(4x+7y).

(12)^3 = 64x^3+343y^3+3×1/21(12).

1728= 64x^3+343y^3+12/7

Now shifting the like terms.

64x^3+343y^3=1728×7/12

64x^3+343y^3=144×7=1008 Ans.

I think u Understand the answer.