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  • Factorize the following : (i)1-64a^{3}-12a+48a^{2}\\ (ii)8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}

Factorize the following :

\\(i)1-64a^{3}-12a+48a^{2}\\ (ii)8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}

 

Answers (1)

(i)(1−4a)(1−4a)(1−4a)

Given, 1−64a3−12a+48a2

The given equation can be written as:

=(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2           ..........(i)

Now we know that:

  a3−b3−3a2b+3ab2=(a−b)3

Comparing equation (i) with the above identity, we get:

∴(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2

=(1−4a)3

Hence the factorized form is (1−4a)(1−4a)(1−4a)

 

(ii)   \left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )

Given, 8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}

The given equation can be written as:

(2p)^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )+\left ( \frac{1}{5} \right )^{3}\\ =(2p)^{3}+\left ( \frac{1}{5} \right )^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )\\                                 …(i)

Now we know that:

a^{3}+b^{3}+3a^{2}b+3ab^{2}=(a+b)^{3}

Comparing equation (i) with the above identity, we get:

(2p)^{3}+\left ( \frac{1}{5} \right )^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )=\left ( 2p+\frac{1}{5} \right )^{3}

Hence the factorized form is\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )         

Posted by

infoexpert24

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