The Expression (A − B)3 + (B − C)3 + (C −A)3 Can Be Factorized as - Mathematics

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MCQ

The expression (a − b)3 + (− c)3 + (c −a)3 can be factorized as

Options

  •  (a − b) (b − c) (c −a)

  •  3(a − b) (b − c) (c −a)

  • −3(a − b) (b −c) (c − a)

  • (a + b + c) (a2 + b2 + c2 − ab − bc − ca)

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Solution

The given expression is

 (a − b)3 + (− c)3 + (c −a)3

Let x= (a-b),  y = (b-c)and z = (c- a). Then the given expression becomes

 (a − b)3 + (− c)3 + (c −a)3 ,` = x^3 + y^3 + z^3`

Note that:

`x+ y+ z = (a-b) + (b-c)+ (c-a)`

                 ` = a-b + b -c + c - a`

                  ` =0`

Recall the formula

  `a^3 +b^3 + c^3 - 3abc = (a+ b+c) (a^2 +b^2 +c^2 - ab - bc - ca)`

When a + b + c =0, this becomes

`a^3 +b^3 +c^3 -3abc = 0.(a^2 +b^2 + c^2 - ab - bc - ca)`

                                   ` = 0`

              ` a^3 +b^3 + c^3 = 3abc`

So, we have the new formula

 ` a^3 +b^3 + c^3 = 3abc` , when a+b+c =0.

Using the above formula, the value of the given expression is

              ` a^3 +b^3 + c^3 = 3abc`

`(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c )(c-a)`

  Is there an error in this question or solution?
Chapter 5: Factorisation of Algebraic Expressions - Exercise 5.6 [Page 25]

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 5 Factorisation of Algebraic Expressions
Exercise 5.6 | Q 5 | Page 25

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