Question

if , 2x+3y=23 and xy=6 find the value of 8x³+27y³​

Answer 1

Answer:

We have:xy=6 so x=6/y

Now putting value of x in the equation 2x+3y=13 then we get,

2*(6/y)+3y=13

12/y+3y=13

(12+3y^2)/y=13

12+3y^2=13y

3y^2–13y+12=0

3y^2–9y-4y+12=0

3y(y-3)-4(y-3)=0

(3y-4)(y-3)=0

So,3y-4=0 or y-3=0

y=4/3 or y=3

Now we have two values of y so we will check which one is correct by relation xy=6,

When y=4/3 ; x=6/y i.e.x=6/(4/3)>>>x=9/2

When y=3; x=6/y i.e.x=6/3>>>x=2

So now by using xy=6,

xy=4/3*9/2>>>2*3>>>6

xy=2*3=6

So both value of x & y satisfy the xy=6 so now will find value of 8x^3+27y^3

8*(9/2)^3+27*(4/3)^3=729+64=793

8*(2)^3+27(3)^3=8*8+27*27=64+729

>>>793

Cubing on both sides

(2x+3y)³=13³

8x³ +27y³+3(2x)(3y)(2x+3y)=13³

According to (a+b)³ formula which is equal to a³+b³+3ab(a+b)

8x³+27y³+18xy(2x+3y)=2197

8x³+27y³+18(6)(13)= 2197

8x³+79327y³+1404=2197

8x³+27y³=2197–1404

== 793

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