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How do you factor x31000 ?

Answer
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Hint: This is a problem related to algebraic identities. This expression given above is of the form a3b3 such that 1000 is the perfect cube of 10. Thus on expanding this identity we will get the factors also. The factors are nothing but the values of x that satisfy the above expression. So we will first expand it and then find the answer.

Complete step-by-step answer:
Given that
 x31000
We can simply write 1000 as a cube of 10.
 x3(10)3
Now we know that a3b3=(ab)(a2+ab+b2) so we can write the above expression as,
 x3(10)3=(x10)(x2+10x+102)
Thus equating this to zero,
 (x10)(x2+10x+102)=0
 (x10)=0
Now taking 10 on right side we get,
 x=10
Now,
 (x2+10x+102)=0
 x2+10x+100=0
Using quadratic formula we get
 b±b24ac2a
Putting the values,
 =10±1024×1×1002×1
On solving the roots,
 =10±1004002
 =10±3002
taking the root value because 100 is a perfect square of 10.
 =10±1032
Taking 10 common and dividing it by 2 we get,
 =5±53i
Thus the roots are
 5+53i&553i
Thus the factors are x=10 and x=5+53i&x=553i .
So, the correct answer is “ x=10 and x=5+53i&x=553i .”.

Note: Note that we can’t directly put the factors as x3=1000 and then 10 is the only possible factor. So we took help from algebraic identities. The factors can be imaginary also. When we find the factors of the second factor that is in the form of algebraic expression their roots can be imaginary depending on the value of discriminant. So since the value of discriminant is negative we get imaginary roots.
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