Question

$\sf{factorise: } \\ \sf \purple{8 ({ x + y})^{3} - 27( {x - y})^{3} }$
___________...
Class IX
Chapter : Factorisation of polynomials...​

Answer 1

Solution :

(-x+5y)(19x²-10xy+7y²)

Step by step explanation :

We have to factorize

$\sf8(x + y)^3-27(x - y)^3$

$\sf=[2(x+y)]^3-[3(x-y)]^3$

$\sf=[2x+2y]^3-[3x-3y)]^3$

We know that $\sf\:a^3-b^3=(a-b)(a^2+ab+b^2)$

$\sf=[2x+2y-(3x-3y)]\times[(2x+2y)^2+(2x+2y)(3x-3y)+(3x-3y)^2]$

$\sf=[-x+5y]\times[4x^2+4y^2+8xy+6x^2-6y^2+9x^2+9y^2-18xy]$

$\sf=[-x+5y]\times[(9+4+6)x^2+xy(8-18)+y^2(4+9-6)]$

$\sf=(-x+5y)(19x^2-10xy+7y^2)$

It is the required solution!

$\rule{200}2$

Algeberaic Indentities :

$\sf\:1)a^3-b^3=(a-b)(a^2+ab+b^2)$

$2)\sf{(a+b)^2=a^2+b^2+2ab}$

$3)\sf{(a-b)^2=a^2+b^2-2ab}$

$4)\sf{(a^2-b^2)=(a+b)(a-b)}$

$5)\sf{(a+b+c)^2={a}^{2} + {b}^{2} + {c}^{2}+2ab+2bc+2ca}$

$6)\sf{(a+b)^3=a^3+b^3+3ab(a+b)}$

Answer 2

Answer:-

To Factorise: 8(x + y)³ - 27(x - y)³

8(x + y)³ - 27(x - y)³

= (2)³(x + y)³ - (3)³(x - y)³

= (2x + 2y)³ - (3x - 3y)³

[∵ xⁿ × yⁿ = (xy)ⁿ]

= [(2x + 2y) - (3x - 3y)] • [(2x + 2y)² + (2x + 2y)(3x - 3y) + (3x - 3y)²]

[∵ a³ - b³ = (a - b)(a² + ab + b²)]

= [2x + 2y - 3x + 3y] • {[(2x)² + 2(2x)(2y) + (2y)²] + [2x(3x - 3y) + 2y(3x - 3y)] + [(3x)² - 2(3x)(3y) + (3y)²]}

= (- x + 5y) • [(4x² + 8xy + 4y²) + (6x² - 6xy + 6xy - 6y²) + (9x² - 18xy + 9y²)]

= (- x + 5y) • (4x² + 8xy + 4y² + 6x² - 6xy + 6xy - 6y² + 9x² - 18xy + 9y²)

= (- x + 5y)(19x² - 10xy + 7y²)

∴ Factorised form of [8(x + y)³ - 27(x - y)³] is [(- x + 5y)(19x² - 10xy + 7y²)].

More:-

Anu, we used the factor finding formula of (a³ - b³) here, i.e. (a - b)(a² + ab + b²). But the value finding formula here is = (a + b)³ - 3ab(a + b).