Question

Factorize: - a³+b³+c³-abc

Answer 1

Answer:

Let f(a)= a3 +b3 +c3 −3abc be a function in a.

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]= a³ + (b+c)³ -3bc(-a) -3abc

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]= a³ + (b+c)³ -3bc(-a) -3abc= a³ + (b + c)³ +3abc -3abc

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]= a³ + (b+c)³ -3bc(-a) -3abc= a³ + (b + c)³ +3abc -3abc[ use, formula , x³ + y³ = (x + y)(x² + y² -xy )

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]= a³ + (b+c)³ -3bc(-a) -3abc= a³ + (b + c)³ +3abc -3abc[ use, formula , x³ + y³ = (x + y)(x² + y² -xy )= {a + (b + c)}{a² + (b+c)² -a(b + c)}

Let f(a)= a3 +b3 +c3 −3abc be a function in a.Now, putting a = -(b+c), we get f( -(b+c)) =b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0Hence, by factor theorem, (a+b+c) is a factor.a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc= a³ + (b +c)³ -3bc(b+c) -3abc[ given, a+ b + c =0 => b+c = -a put it ]= a³ + (b+c)³ -3bc(-a) -3abc= a³ + (b + c)³ +3abc -3abc[ use, formula , x³ + y³ = (x + y)(x² + y² -xy )= {a + (b + c)}{a² + (b+c)² -a(b + c)}=(a + b + c)(a² + b² + c² + 2bc – ab – ac)

Step-by-step explanation:

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