Question

plz help to factorise it....

Answer 1

Answer:

Hope that Helps.....

Step-by-step explanation:

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Answer 2

Answer:

(√2a+2b-3c) (2a²+4b²+9c²-2√2 ab+6bc+3√2ac)

Step-by-step explanation:

let z=2√2a³+8b³-27c³+18√2abc

z=(√2a)³+(2b)³+(-3c)³-3*(√2a*2b)(-3c)

Let √2a=p,2b=q, -3c=r

z=p³+q³+r³-3pqr

=(p+q+r)(p²+q²+r²-pq-qr-pr)

putting values of p,q,r

z=(√2a+2b-3c) { (√2a)²+(2b)²+(-3c)²-(√2a)(2b)-(2b)(-3c)-(-3c)(√2a) ]

=(√2a+2b-3c) (2a²+4b²+9c²-2√2 ab+6bc+3√2ac)