Question

if 3 A + 5b + 4 C = 0 , prove that: 27 a^ cube + 125 b ^cube + 64 C^ cube equals to 180 ABC​

Answer 1

By using the formula ,

a3+ b3+ c3= 3abc, where a+b+c= 0.

then,

(3a)3+ (5b)3 +(4c)3 = 3*3a*5b*4c

= 27a3 + 125b3+ 64c3= 180abc

Hence proved.

Answer 2

Here we use Identity  :

If $x+y+z=0$ , then $x^3+y^3+z^3=3xyz$  

Step-by-step explanation:

Identity in polynomials :

If $x+y+z=0$ , then $x^3+y^3+z^3=3xyz$   (1)

To prove : $27 a^3 + 125 b ^3 + 64 c^3=180abc$

Consider L.H.S. $27 a^3 + 125 b ^3 + 64 c^3$

= $=3^3a^3+5^3b^3+4^3c^3$

$= (3a)^2+(5b)^3+(4c)^3$

Since , it is given that $3 a + 5b + 4 c = 0$ m, so by using (1) , we get

$= (3a)^2+(5b)^3+(4c)^3= 3(3a)(5b)(4c)= 180abc$  = R.H.S.

Hence, proved.

# Learn more :

Prove that (a+b)cube+(b+c)cube+(c+a)cube-3(a+b)×(b+c)×(c+a)=2(acube+bcube+ccube)

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