Question

If A+ 2B + 3C = 0, prove that A³ + 8B
³ + 27C³ = 18ABC.​

Answer 1

Answer:

Step-by-step explanation:

if a+b+c=0

$a^{3} +b^{3} +c^{3}=3abc\\ a=A\\b=2B\\c=3C\\then\\A^{3}+(2B)^{3} +(3C)^{3} =A³ + 8B³ + 27C³=3*A*2B*3C=18ABC..$

PLS MARK AS BRAINLIEST

Answer 2

Answer:

Step-by-step explanation:

To prove A^3 + 8B^3 + 27C^3 = 18ABC

PROOF

Step 1 = As we know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

So, Similarly

A^3 + 8B^3 + 27C^3 - 3(A)(2B)(3c) = (A + B + C)(A^2 + B^2 + C^2 - ab - bc - ca)[using identity]

According to the question A + 2B + 3C = 0(given)

L.H.S = A^3 + 8B^3 + 27C^3 - 3 X 2 X 3 X A X B X C

        = A^3 + 8B^3 + 27C^3 - 18ABC

R.H.S =( A + B + C)(A^2 + B^2 + C^2 - ab - bc - ca)

A = A, B = 2B, C = 3C

         =( A + 2B + 3C)(A^2 + 2B whole^2 + 3C whole^2 - (A)(2B) - (2B)(3C) - (3C)(A)[using identity]

According to the question, A + 2B + 3C = 0

And as we know anything multiplied by 0 is 0

          = (0)(A^2 + 4B^2 + 9C^2 - 2AB - 6BC - 3CA)

R.H.S  = 0

Step 2 = To prove A^3 + 8B^3 + 27C^3 = 18ABC

we have to equate L.H.S and R.H.S

That is,

L.H.S = R.H.S

A^3 + 8B^3 + 27C^3 - 18ABC = 0

A^3 + 8B^3 + 27C^3 = 18ABC(We will take the 18ABC from the L.H.S by following the linear equation rules and will keep it in the R.H.S by changing the sign[A^3 + B^3 + C^3 = 3ABC])

HENCE PROVED