Question

Factorise the following:
x²-(x-y)⁴​

Answer 1

Solution↷

${x}^{2} - (x - y) {}^{4} \\ = [ {x}^{2} + (x - y) {}^{2} ][ {x}^{2} - (x - y) {}^{2} ] \\ = ( {x}^{2} + {x}^{2} - 2xy + {y}^{2})(x + (x - y) \\ = (2xy {}^{2} - 2xy + {y}^{2} )(2x - y)(x - x + y) \\ \green{ = y(2x - y)(2 {x}^{2} - 2xy + {y}^{2}) }$

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Factorize:

• x² - (x - y)⁴

Solution:

$x^2 - (x - y)^4 \\\\\implies x^2 - \left((x - y)^2\right)^2 \\\\\text{As, a^2 - b^2 = (a + b)(a -b)} \\\\\implies \left(x^2 + (x - y)^2\right) \times \left(x^2 - (x - y)^2\right) \\\\\implies \left(x^2 + (x^2 + y^2 - 2xy)\right) \times \left(x^2 - (x^2 + y^2 - 2xy) \right) \\\\\implies \left(x^2 + x^2 + y^2 - 2xy\right) \times \left(x^2 - x^2 - y^2 + 2xy\right) \\\\\implies \left(2x^2 + y^2 - 2xy\right) \times \left(2xy - y^2\right) \\\\\implies \left(2x^2 + y^2 - 2xy\right) \times \left(y(2x - y)\right) \\\\\bf{\implies y * (2x - y) * (2x^2 - 2xy + y^2)}$

Formula Used:

• $a^2 - b^2 = (a + b)(a -b)$
• $(a - b)^2 = a^2 + b^2 - 2ab$

Hope it helps!!