hey anyone can solve my problem pleaseeeee
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Given :
(a + b + c)^3 - a^3 - b^3 - c^3: - -------- (1)
[(a + b) + c)^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3
= a^3 + 3a^2b + 3ab^2 + b^3 + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2 + c^3
= a^3 + 3a^2b + 3ab^2 + b^3 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3 ----- (2)
Now,
Substiute (2) in (1), we get
= > a^3 + 3a^2b + 3ab^2 + b^3 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3 - a^3 - b^3 - c^3
= > 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2
= > 3(a^2b + ab^2 + a^2c + 2abc + b^2c + ac^2 + bc^2)
= > 3(ab + ac + b^2 + bc)(c + a)
= > 3(a + b)(b + c)(c + a).
Hope this helps!
(a + b + c)^3 - a^3 - b^3 - c^3: - -------- (1)
[(a + b) + c)^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3
= a^3 + 3a^2b + 3ab^2 + b^3 + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2 + c^3
= a^3 + 3a^2b + 3ab^2 + b^3 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3 ----- (2)
Now,
Substiute (2) in (1), we get
= > a^3 + 3a^2b + 3ab^2 + b^3 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3 - a^3 - b^3 - c^3
= > 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2
= > 3(a^2b + ab^2 + a^2c + 2abc + b^2c + ac^2 + bc^2)
= > 3(ab + ac + b^2 + bc)(c + a)
= > 3(a + b)(b + c)(c + a).
Hope this helps!
siddhartharao77:
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