Question

PTDT :
(a+b+c)³ - a³-b³-c³

Answer 1

(a + b + c)³ - a³ - b³ - c³

= (a+b)³ + 3(a+b)c² +3(a+b)²c+c³- a³ - b³ - c³

= a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3(a+b)c²+c³- a³ - b³ - c³

= 3a²b+3ab³+3a²c+6abc+3b²c+3c²a+3c²b

= (3c²b+6abc+3a²b)+3ac²+(3ab²+3b²c)+3a²c

= 3b(c²+2ac+a²)+3ac²+3b²(a+c)+3a²c

= 3b(c+a)²+3ac(a+c)+3b²(a+c)

= 3(a+c)[b(a+c)+ac(a+c)+b²]

Answer 2

(a+b+c)³ - a³-b³-c³

= a³ +b³ +c³ + 3(a+b)(a+c)(b+c) - a³ - b³ - c³

= 3(a+b)(a+c)(b+c)

Answer 3

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