Question

if 2x- 3y=10and xy=16.Find of 8x^3-27y^3​

Answer 1

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• 2x-3y=10
• xy=16

${\sf{\green{\underline{\large{To\:find}}}}}$

• 8x³-27y³

${\sf{\pink{\underline{\Large{Explanation}}}}}$

We know that a³+b³

=>a³+b³ = (a+b) (a²+b²-ab)

Breaking 8x³-27y³

(2x)³-(3y)³

=>(2x)³-(3y)³=(2x-3y) {(2x-3y)²-6xy)}

¶utting the values

___________________________

=>(2x)³-(3y)³=(2x-3y) {(2x-3y)²-6xy)}

=>(2x)³-(3y)³=(10) {(10)²-6×16)}

=>(2x)³-(3y)³=(10) {(100)-96)}

=>(2x)³-(3y)³=(10) {4)}

=>8x³-27y³=40

___________________________

Answer 2

Step-by-step explanation:

2x-3y=10

xy=16

x=16/y

2 (16/y)-3y=10

32/y-3y=10

32-3y^2/y=10

32-3y^2=10y

-3y^2-10y+32=0

3y^2+10y-32=0

by using qudratic formula we can find out the value of y

y=2

x=16/2

x=8

8x^3-27y^3

8 (8)^3-27 (2)^3

4096-216

3880