Math, asked by vanshkanojia359, 5 months ago

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

Attachments:

Answers

Answered by Cynefin
30

 \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)
Answered by Anonymous
117

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}}

Similar questions