Question

if 2a+ 3b +c = 0 then show that 8a³ + 27 b³ + c³ =18 abc​

Answer 1

Given:

✰ 2a+ 3b + c = 0

To Show:

✠ 8a³ + 27 b³ + c³ = 18 abc

Proof:

We have,

2a+ 3b + c = 0

Transpose c,

2a + 3b = - c ...①

Cubing both the sides, we get

➤ ( 2a + 3b )³ = ( - c )³

Use identity,

(a + b)³ = ( a³ + b³ + 3ab ( a + b )

➤ ( (2a)³ + (3b)³ + 3 × 2a × 3b ( 2a + 3b ) = -c³

• Now, first do cube of coefficient and then variable. 2 × 2 × 2 = 8 and a³. Same for 2nd term - 3 × 3 × 3 = 27 and b³. Now, multiply constant with coefficients first ( 3 × 2 × 3 = 18 ), now multiply both the variables ( a × b = ab ), At last combine them all, we get:

➤ 8a³ + 27b³ + 18ab ( 2a + 3b) = -c³ ...②

Substitute the value of 2a + 3b from eq①, we have

➤ 8a³ + 27b³ + 18ab ( - c ) = -c³

➤ 8a³ + 27b³ - 18abc = -c³

• On transfering any term, to left hand side sign will change, therefore On transfering c to to left hand side sign will change.

➤ 8a³ + 27b³ + c³ = + 18abc

➤ 8a³ + 27b³ + c³ = 18abc

∴ 8a³ + 27b³ + c³ = 18abc

Hence Proved!!

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