Question

if x plus 3y equals 6 then prove that x cube plus 27y cube plus 54 xy equals 216​

Answer 1

Answer:

$x + 3y = 6$

squaring both the sides

$(x + 3y)^{2} = (6)^{2}\\\\x^{2} + (3y)^{2} + 2(x)(3y) = 36\\\\x^{2} + 9y^{2} + 6xy = 36\\\\x^{2} + 9y^{2} = 36 - 6xy$

To prove:

$x^{3} + 27y^{3} + 54xy = 216$

$x^{3} + (3y)^{3} + 54xy = 216\\\\(x + 3y)[x^{2} + (3y)^{2} - (x)(3y)] + 54xy = 216\\\\(6)(36 - 6xy - 3xy) + 54xy = 216\\\\(6)(36 - 9xy) + 54xy = 216\\\\216 - 54xy + 54xy = 216\\\\216 = 2160$

Answer 2

SOLUTION :

x + 3y = 6

Cubing both sides

x³ + (3y)³ + 3 × x × 3y (x + 3y) = 6³

x³ + 27y³ + 3 × x × 3y ×6 = 6³

x³ + 27y³ + 54xy = 216

Hence the proof follows