Question

Find the value of x^3-27y^3-343-27xy when x=3y+7

Answer 1

Answer:

$108y^{2}+252y$

Step-by-step explanation:

The given equation is:

$x^{3}-27y^{3}-343-27xy$

Putting x= 3y+7 in the above equation, we have

$(3y+7)^{3}-27y^{3}-343-27(3y+7)y$

Using $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$, we have

=$27y^{3}+343+63y(3y+7)-27y^{3}-343-27(3y^{2}+7y)$

=$189y^{2}+441y-81y^{2}-189y$

=$108y^{2}+252y$

which is the required value.

Answer 2

Step-by-step explanation:

125x^3-27y^3-225xy^2

(5x)^3-(3y)^3-45xy(5x-3y)

=(5x)^3-(3y)^3-3×5x+3y(5x-3y)

=(5x-3y)^3

=(5x-3y)(5x-3y)(5x-3y)

I think it is correct

if it is wrong please forgive me