Advertisements
Advertisements
प्रश्न
The expression (a − b)3 + (b − c)3 + (c −a)3 can be factorized as
विकल्प
(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)
Advertisements
उत्तर
The given expression is
(a − b)3 + (b − c)3 + (c −a)3
Let x= (a-b), y = (b-c)and z = (c- a). Then the given expression becomes
(a − b)3 + (b − 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)`
APPEARS IN
संबंधित प्रश्न
Factorize xy9 - yx9
Factorize `2x^2 - 5/6x + 1/12`
Find the value of the following expression: 16x2 + 24x + 9, when \[x = \frac{7}{4}\]
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :
8 − 3x, xy2, 3y2 − 5y + 8, 9x − 3x2 + 15x3 − 7,
3x × 5y, 3x ÷ 5y, 2y ÷ 7 + 3x − 7 and 4 − ax2 + bx + y
Evaluate: (a2 + 2ab + b2)(a + b)
Divide: 2m3n5 by - mn
Divide: 5x2 - 3x by x
Write the coefficient of x2 and x in the following polynomials
πx2 – x + 2
If x = 2 and y = 3, then find the value of the following expressions
x + 1 – y
3p2 – 5pq + 2q2 + 6pq – q2 + pq is a
