Question

factorise 27p^3 + 108p^2q + 144pq^2 + 64q^3​

Answer 1

Answer:

here your answer are given below

Step-by-step explanation:

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

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

Since,

3p(4q - 2r) + 4q(2r - 3p) + 2r(3p - 4q)

= 12pq - 6pr + 8qr - 12pq + 6pr - 8qr = 0

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

= 3×3p(4q - 2r) + 4q(2r - 3p) + 2r(3p - 4q)

= 72pqr(3p - 4q) (4q - 2r) (2r - 3p)

Answer.