please solve this fastas possible
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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 ^_^
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:
:-)
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