Question

factorise 27p³ -1/216 - 9p²/2+ 1p/4

Answer 1

27p³(4q-2r)³+64q³(2r-3p)³+8r³(3p-4q)³ =27p³(64q³-3.16q².2r+3.4q.4r²-8r³)+64q³(8r³-3.4r².3p+3.2r.9p²-27p³)+8r³(27p³ -3.9p².4q+3.3p.16q²-64q³) =1728p³q³-2592p³q²r+1296p³qr²-216p³r³+512q³r³-2304pq³r²+3456p²q³r1728p³q³+216p³r³-864p²qr³+1152pq²r³-512q³r³ =1296p³qr²+3456p²q³r+1152pq²r³-2592p³q²r-2304pq³r²-864p²qr³ =48pqr(27p²r+72pq²+24qr²-54p²q-48q²r-18pr²)
=48pqr[(27p²r-18pr²)+(24qr²-54p²q)+(72pq²-48q²r)]
=48pqr[9pr(3p-2r)-6q(9p²-4r²)+24q²(3p-2r)] =48pqr[9pr(3p-2r)-6q(3p+2r)(3p-2r)+24q²(3p-2r)]
=48pqr(3p-2r)[9pr-6q(3p+2r)+24q²]

Factirize 27p 3 (4q 2r) 3 +64q 3 (2r-3p) 3 +8r 3 (3p-4q) 3
=48pqr(3p-2r)(9pr-18pq-12qr+24q²)
=48pqr(3p-2r)[9p(r-2q)-12q(r-2q)]
=48pqr(3p-2r)(r-2q)(9p-12q)
=48pqr{3(3p-4q)}(3p-2r)(r-2q)
=144pqr(3p-4q)(3p-2r)(r-2q)

Answer 2

Step-by-step explanation:

Given:-

$\tt{27 {p}^{3} - \frac{1}{216} - \frac{9}{2} {p}^{2} + \frac{1}{4} p}$

What to do:-

To Factorise the expression.

Solution:-

Let's solve the problem,

We have,

$\tt{27 {p}^{3} - \frac{1}{216} - \frac{9}{2} {p}^{2} + \frac{1}{4} p}$

$⟹ \tt{(3p {)}^{2} - \bigg( \frac{1}{6} \bigg )^{2} - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }$

$⟹ \tt{\bigg(3p - \frac{1}{6 }\bigg)\left\{(3p {)}^{2} + 3p \times \frac{1}{6} + \bigg({ \frac{1}{6} \bigg)^{2} }\right\} - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }\\ \\ </p><p>$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} + p \times \frac{1}{2} + \frac{1}{36} \bigg) - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} + \frac{p}{2} + \frac{1}{36} - \frac{3}{2} p \bigg)}$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} - p + \frac{1}{36} \bigg)}$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \left \{(3p {)}^{2} - 2 \times 3p \times \frac{1}{6} + \bigg( \frac{1}{6} \bigg)^{2} \right \} }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg )^{2} }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg) } \: \: \red{Ans}.$

I hope it's help you...☺