please solve all Question
factorise ( given in the picture)
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here's the answer
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hope it helps you
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Step-by-step explanation:
28. (a+b-c)³ + (a-b+c)³ - 8a³
= (a+b-c)³ + (a-b+c)³ + (-2a)³
= 3.(a+b-c).(a-b+c).(-2a)
[ if, (a+b+c)=0. then, a³+b³+c³=3abc]
= -6a.[{a+(b-c)}.{a-(b-c)}]
= -6a.[a²- b²+2bc-c²]
= -6a³+6ab²-12abc+6ac²
= 6ab²+6ac²-6a³-12abc
Equally the others sums will be solved
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