Question

$(81a { }^{2} - 4) \div (9a + 2)$
​

Answer 1

ANSWER:

To Solve:

• (81a² - 4) ÷ (9a + 2)

Solution:

$\text{We are given that,}\\\\:\implies\dfrac{81a^2-4}{9a+2}\\\\\text{We know that,}\\\\:\longrightarrow81=9\times9=9^2\\\\\text{And,}\\\\:\longrightarrow4=2\times2=2^2\\\\\text{So,}\\\\\:\implies\dfrac{81a^2-4}{9a+2}\\\\:\implies\dfrac{9^2a^2-2^2}{9a+2}\\\\:\implies\dfrac{(9a)^2-2^2}{9a+2}\\\\\text{We know that,}\\\\:\longrightarrow a^2-b^2=(a+b)(a-b)\\\\\text{So,}\\\\:\implies\dfrac{(9a)^2-2^2}{9a+2}\\\\:\implies\dfrac{(9a+2)(9a-2)}{9a+2}\\\\\text{(9a+2) gets cut,}\\\\\text{So,}\\\\\bf{:\implies 9a-2}$

Formula Used:

• a² - b² = (a + b)(a - b)

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$