Question

selamm...

quick and short answer

*PYBS* MY HAVE ONLINE EXAMPLES

I need to work urgently


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Answer 1

The right option is E.

Step-by-step explanation:

To Solve:

$\dfrac{( {a}^{x} + a {}^{y} )( {a}^{2x} - {a}^{x + y} + {a}^{2y}) - {a}^{3x} }{{a}^{y}}$

Ne yapmalı?

This question is both mixture of Laws of Indices and Algebra. So we will solve it by using these rules.

1st) Split the terms in Simple forms like,

$a^{c+b} = a^c.a^b$

2nd)Use Formula as required.

a³ + b³ = (a+b)(a²-ab+b²) will be used.

Solution:

$\dfrac{( {a}^{x} + a {}^{y} )( {a}^{2x} - {a}^{x + y} + {a}^{2y}) - {a}^{3x} }{{a}^{y}} \\ \\= \dfrac{( {a}^{x} + a {}^{y} )( {a}^{2x} - {a}^{x}. {a}^{y} + {a}^{2y})- {a}^{3x}}{{a}^{y}} \\ \\ = \frac{ {a}^{3x} + {a}^{3y}-{a}^{3x} }{ {a}^{y} } \\= \frac{ {a}^{3y} }{ {a}^{y} } \\ \\= {a}^{(3y - y)} \\ \\ = {a}^{2y}$

Therefore, the required answer is $\boxed{{a}^{2y}}$

${\boxed{\pink{\text{ Some Important Laws of Indices}}}}$

${a}^{n}.{a}^{m}={a}^{(n + m)}$

${a}^{-1}=\dfrac{1}{a}$

$\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}$

${({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}$

${a}^{\frac{1}{x}}=\sqrt[x]{a}$

[Where all variables are real and greater than 0]

Answer 2

Answer:

answer option e-a^2y

hope it helps you..

thankyou..