Math, asked by Rahika1, 1 year ago

please solve this fastas possible

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Answered by Anonymous
4
Hi there !!

Given,

To factorise,

27p³(4q-2r)³ + 64q³(2r-3p)³ + 8r³(3p-4q)³

The following expression is in the form of the algebraic identity

a³ + b³ + c³ = 3abc

Where,
a = 3p(4q - 2r ) since (3p(4q-2r)³ = 27p³(4p-2r)³

b = 4q(2r - 3p) since (4p(2r - 3p) )³=64q³(2r-3p)³

c = 2r(3p-4q) since (2r(3p-4q) )³ = 8r³(3p-4q)³

So,

3[3p(4q-2r)(4q(2r-3p)(2r(3p-4q)]

= 3( 12pq - 6pr)( 8qr - 12pq)(6pr - 8qr) is the answer


Hope it helped you ^_^

Anonymous: :-)
Rahika1: is it to be factorize more and are yu sure about the answer
Rahika1: please. give correct answer
Rahika1: and are you sure
Anonymous: yep, i am sure about my answer
Rahika1: Kya Tumne Kisi Se Dekha
Anonymous: nope , but i am sure
Anonymous: thanks for the brainliest :D
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