factorize (x+1)3-(x-1)3
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0
do multiply those brackets with outer number...
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3
Step-by-step explanation:
Here is
=> (x+1)³ - (x-1)³
[ (a+b)³ = a³+3a²b+3ab²+b³ ]
[ (a-b)³ = a³-3a²b+3ab²-b³ ]
=> [ x³+3(x)²(1)+3(x)(1)²+(1)³ ] - [ x³-3(x)²(1)+3(x)(1)²-(1)³]
=> x³+3x²+3x+1 - x³+3x²-3x+1
=> 6x²+2
=> 2(3x²+1)
Hope it hlpz...
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