Try WA, I double dare you!
3 8 + x + 3 8 − x = 1
Find the product of the root(s) of the equation above.
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4 solutions
Use L A T E X for your equations, it would help in communication of ideas. I had no idea what your question was asking until I saw the solution. Your problem has been edited btw.
Exactly same way
If a + b − c = 0 , then a 3 + b 3 − c 3 = 3 a b ( − c ) .
We have, 3 8 + x + 3 8 − x − 1 = 0
⟹ ( 3 8 + x ) 3 + ( 3 8 − x ) 3 − ( 1 ) 3 = 3 ( 3 8 + x ) ( 3 8 − x ) ( − 1 )
After simplifying we get,
− 5 = 3 6 4 − x 2
Cubing both sides,
x 2 = 1 8 9
Hence product of roots = − 1 8 9
This is actually a better solution than the other.
However, to generalize the formula you applied, it is better to write
For real number a , b , c , if a + b + c = 0 , we have a 3 + b 3 + c 3 = 3 a b c
than
If a + b − c = 0 , then a 3 + b 3 − c 3 = 3 a b ( − c )
so that others can know more about it, rather than just a specific case.
By the way,
a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a )
Actually I used the same method as Aditya Chauhan, but I did try WA and it yielded the same results:
Hey guys.I have a doubt...might sound silly but try substituting the value of x ,if u substitute positive value of x ,u get the second term imaginary.and if u substitute negative value of x u get first term imaginary. And although the vice versa part happens to be real..I really don't understand how can a real number and an imaginary number add up to 1? .any guidance is appreciated.thanks..
Clearly x ∈ [ − 8 , 8 ] .
So you can't make any random substitution and substituting any value of x between − 8 and 8 won't give you imaginary numbers.
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Wel when x^2 =189…,x is either root 189 or -ve root 189. So though the limits of x seem to be extending between [-8,8] the actual values of x don't fit in .so I guess the answer should have been no solutions exist.
Power is 'odd' root so they won't yield any imaginary number for example
( − 8 ) 3 1 = − 2
You will get imaginary result only when there in ' even ' root.
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Why?
Dont' we have ( − 1 ) 1 / 3 = − 1 , − e 2 i π / 3 , − e 4 i π / 3
So why are we considering only the real roots?
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Shit! I forgot imaginary roots -_-
Because both terms have different imaginary part?
Cubing the eq. ,we get:
8 + x + 8 − x + 3 3 ( 8 + x ) ( 8 − x ) ( 3 8 + x + 3 8 − x ) = 1 1 6 + 3 3 6 4 − x 2 = 1 3 3 6 4 − x 2 = − 1 5 3 6 4 − x 2 = − 5
Cubing again, we get:
6 4 − x 2 = − 1 2 5
x 2 − 1 8 9 = 0
Therefore product of roots=-189