Question

If
2x + 3y = 12
12 and
xy = 6 , find the value of 8x cube+27y cube​

Answer 1

Given :-

2x+3y = 12 》equation.1

xy = 6 》equation.2

To Find :-

$= > 8x {}^{3} + 27y {}^{3}$

Solution :-

From the equation.2 》

xy = 6

y = 6/x

"y" value substitute in equation.1》

2x+3(6/x) = 12

2x+(18/x) = 12

(2x × x + 18)/x = 12

2x^2 +18 = 12x

2x^2-12x+18 = 0 》equation.3

》We have to find the "x" value by using finding roots method.

2x^2-6x-6x+18 = 0

2x(x-3)-6(x-3) = 0

(2x-6)(x-3) = 0

2x-6 = 0 or x-3 = 0

2x = 6 or x = 3

x = 3 or x = 3

soo we got 》 x = 3

》"x" value substite in equation.2 then we get "y" value.

xy = 6

3y = 6

y = 6/3

y = 2

》 Finally we have to find the value of

"8x^3+27y^3 " by using "x & y" values.

=> 8(3)^3 + 27(2)^3

=> 8(27) + 27(8)

=> 216 + 216

=> 432

Checking the "x & y" values :-

》 take the equation.1 and substitute the values of "x & y".

2x + 3y = 12

2(3) + 3(2) = 12

6 + 6 = 12

12 = 12

LHS = RHS

Answer 2

2x+3y=12

Xy =6

Find 8x^3+27y^3

2x+3y=12

Cubing both sides

(2x+3y)^3=12^3

(a+b)^3=a^3+b^3+3ab(a+b)=1728

8x^3+27y^3+3×2x×3y(2x+3y)=1728

8x^3+27y^3+18×6×12=1728

8x^3+27y^3+1296=1728

8x^3+27y^3= 432

i hope it will helps u