Question

(2p+3p)(4p²-6pq+9q²)​

Answer 1

${\large{\pmb{\sf{\underline{\bigstar \: \: Required \: Solution...}}}}}$

~ This question says that we have to expand the following by using suitable identity. Let us expand the given equation by using identity.

We are asked to expand:

$\: \: \: \:{\small{\underline{\boxed{\sf{(2p+3q)(4p^{2}-6pq+9q^{2})}}}}}$

Suitable identity:

$\: \: \: \:{\small{\underline{\boxed{\sf{a^{3} + b^{3} \: = (a+b)(a^{2}-ab+b^{2})}}}}}$

Full Solution:

${\sf{\dashrightarrow 2p+3q)(4p^{2}-6pq+9q^{2})}}$

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

${\sf{\dashrightarrow (2p^{3} + 3q^{3})}}$

$\: \: \: \:{\small{\underline{\boxed{\sf{Solution \: = (2p^{3} + 3q^{3})}}}}}$

Additional information:

• Some identities:

$\; \; \; \; \; \; \;{\sf{\leadsto (A+B)^{2} \: = \: = A^{2} \: + \: 2AB \: + B^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (A-B)^{2} \: = \: = A^{2} \: - \: 2AB \: + B^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto A^{2} \: - B^{2} \: = \: (A+B) \: (A-B)}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (A+B)^{2} \: = (A-B)^{2} \: +4AB}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (A-B)^{2} \: = (A+B)^{2} \: -4AB}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (A+B)^{3} \: = A^{3} + \: 3AB \: (A+B) \:+ B^{3}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (A-B)^{3} \: = A^{3} - \: 3AB \: (A-B) \: + B^{3}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto A^{3} \: + B^{3} = \: (A+B) (A^{2} - AB + B^{2}}}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$

• Knowledge about Quadratic equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━

Answer 2

__________________

♧Question♧

• (2p+3p)(4p²-6pq+9q²)

♧Answer♧

$\dashrightarrow\rm{(2p+3p)(4p²-6pq+9q²)}$

By using this formula,

$\rm \red{∵(a^{3} + b^{3}) = (a + b)(a^{2} - ab + b^{2})}$

☆So we can write it as.

$\dashrightarrow \boxed {\red{ \rm{(2p^{3} +3p^{3} )}}} \: \: \: \: \: \large\rm{Ans}$

Additional information:-

• ➠(a³ + b³) = (a + b)(a² - ab + b²)
• ➠(a³ - b³) = (a - b)(a² + ab + b²)
• ➠(a + b)² - (a - b)² = 4ab
• ➠2(a² + b²) = (a + b)² + (a - b)²

__________________