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Question

Solve the following cubic equation :

x3+x216x=16

A
x=1,±4
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B
x=1,±4
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C
x=0,±3
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D
none of these
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Solution

The correct option is A x=1,±4
x3+x216x=16

x3+x216x16=0

Let a(x)=x3+x216x16

a(1)=1+1+1616=0

Therefore, a(x)=(x+1)(x216)

0=(x+1)(x+4)(x4)

x=1,4,4

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