Question

using identity 1/3×+2/3y) (1/3×-2/3y)​

Answer 1

$\huge\bf{\red{\underline{\underline{\mathcal{AnSwer}}}}}$

$\sf\dfrac{1}{9}x² - \sf\dfrac{4}{9}y²$

Identity:-

$\underline{\boxed{\sf (a+b)(a-b) = a²-b² }}$

Solution:-

$\longrightarrow \bigg(\sf \dfrac{1}{3}x + \dfrac {2}{3}y\bigg)\bigg(\sf \dfrac{1}{3}x - \dfrac {2}{3}y\bigg)$

$\longrightarrow {\bigg(\sf\dfrac{1}{3}x\bigg)}^{2} - {\bigg(\sf\dfrac{2}{3}y\bigg)}^{2}$

$\longrightarrow\sf\dfrac{1}{9}x² - \sf\dfrac{4}{9}y²$

_______________________

Additional information:-

$\longrightarrow$ (a+b)² = a²+b²+2ab

$\longrightarrow$ (a-b)² = a²+b²-2ab

Answer 2

$\bf{ \underline{Given:- }}$

$( \frac{1}{3}x + \frac{2}{3} y)( \frac{1}{3} x - \frac{2}{3} y)$

$\huge\bf{ \underline{Solution :- }}$

$\sf{We \: know \: that, }$

$\bf \red{• \: (a+b)(a-b) = {a}^{2 } - {b}^{2} }$

$\sf{So, }$

$\bf( \frac{1}{3}x + \frac{2}{3} y)( \frac{1}{3} x - \frac{2}{3} y)$

$\rightarrow \bf (\frac{1}{3} x )^{2} - ( \frac{2}{3} y) ^{2}$

$\rightarrow \bf \frac{1}{9} {x}^{2} - \frac{4}{9} {y}^{2}$

$\bf \pink {Answer: \frac{1}{9} {x}^{2} - \frac{4}{9} {y}^{2} }$

$\bf \purple{ \underline{Important \: Informations:-}}$

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

$• \: ( {a - b})^{2} = {a}^{2} - 2ab + {b}^{2}$

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

$• \: 4ab = {(a + b)}^{2} - {(a - b)}^{2}$

$• \: {(a + b)}^{3} = {a}^{3} + 2 {a}^{2} b + 2a {b}^{2} + {b}^{3}$

$• \: {(a - b)}^{3} = {a}^{3} - 2 {a}^{2} b + 2a {b}^{2} - {b}^{3}$

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

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