if p= 2a' then prove that a3+6a p+ p3- 8 = 0
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Hey I am vipul here
i hope this answer may helpful for u
there is a fault in questions
it should p=2-a
p=2-a
cube on both side
then we find
p3={2-a}3
p3=23-a3-3.2.a{2-a}
and {2-a}=p
p3=8-a3-6ap
p3-8+a3+6ap=0
a3+6ap+p3-8=0
thank u
i hope this answer may helpful for u
there is a fault in questions
it should p=2-a
p=2-a
cube on both side
then we find
p3={2-a}3
p3=23-a3-3.2.a{2-a}
and {2-a}=p
p3=8-a3-6ap
p3-8+a3+6ap=0
a3+6ap+p3-8=0
thank u
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