if p=2-a , prove that a cube+6ap+p cube -8=0
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if p =2-a
then a*a*a + 6*a*(2-a) + (2-a)*(2-a)*(2-a)-8 =0
a cube + 12a-(6a*a) + 8-12a+(6a*a) + a cube - 8=0
a*a*a + a*a*a = 0
2a cube = 0
a cube =0
a= 0
now substitute the value of a and it will give you 0
then a*a*a + 6*a*(2-a) + (2-a)*(2-a)*(2-a)-8 =0
a cube + 12a-(6a*a) + 8-12a+(6a*a) + a cube - 8=0
a*a*a + a*a*a = 0
2a cube = 0
a cube =0
a= 0
now substitute the value of a and it will give you 0
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