Question

if (x+3y)=6 & xy=2,find the value of (x^3+27y^3)​

Answer 1

✏ Answer

$(x + 3y) {}^{3} = x {}^{3} + 27y {}^{3} + 3(xy3)(x + 3y) \\ (6) {}^{3} = x {}^{3} + 27y { }^{2} + 3 \times 3(2)(6) \\ 216 = {x}^{3} + 27y {}^{3} + 108 \\ 216 - 108 = x {}^{3} + 27y {}^{3} \\ x {}^{3} + 27y {}^{3} = 108$

Answer 2

Answer:

Step-by-step explanation:

you know x cube + (3y) cube = x cube + 27y cube

and x cube + (3y) cube = (x+3y) cybe - 3xy(x+3y)

                                     = 6 cube - 6 x 6

                                       = 216 - 36

                                     = 180