Question

please tell me someone and explain me please I will mark you as brainlist promise and it should be step by step please ​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

To Factorise:

• $\rm{ \dfrac{ {x}^{3}y }{9} - \dfrac{x {y}^{3} }{16} }$

Step-by-step Explanation:

Taking xy common from both terms:

$\rm{xy \bigg( \dfrac{ {x}^{2} }{9} - \dfrac{ {y}^{2} }{16} \bigg)}$

Now have a look inside the parentheses. Both the terms are squares. They can be written as,

$\rm{xy \bigg(( \dfrac{x}{3} ) {}^{2} - ( \dfrac{y}{4} ) {}^{2} } \bigg)$

We can use the identity a² - b² = (a+b)(a-b),

$\rm{xy \bigg( \dfrac{x}{3} + \dfrac{y}{4} \bigg) \bigg( \dfrac{x}{3} - \dfrac{y}{4} \bigg)}$

And we are done! :D

Explore more!!

• Identities can be used for factorising polynomials with higher degrees.

The Algebraic Identities are:

• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• a² – b²= (a + b)(a – b)
• (x + a)(x + b) = x² + (a + b) x + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + b³ + 3ab (a + b)
• (a – b)³ = a³ – b³ – 3ab (a – b)
• a³ + b³ + c³ – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Answer 2

Given :

• x³y/9 - xy²/ 16

To Find :

• Find the factories

Solution :

$\sf : \implies \: \: \: \frac{ {x}^{3} y}{9} - \frac{ x {y}^{3} }{16} \\ \\ \\ \sf : \implies \: \: \: xy \bigg( \frac{ {x}^{2} }{9} - \frac{ {y}^{2} }{16} \bigg) \\ \\ \\ \sf : \implies \: \: \: xy\bigg( ({\frac{x}{3}})^{2} - ({\frac{y}{4}})^{2} \bigg)$

We use :

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

Substitute all values :

$\sf : \implies \: \: \: xy\bigg( {\frac{x}{3} + {\frac{y}4}} \bigg) \bigg( {\frac{x}{3} - {\frac{y}4}} \bigg)$

More to know:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$