Cubic Polynomial
Let p ( x ) and q ( x ) be two polynomials, both of which have their sum of coefficients equal to s > 0 . Let p and q satisfy p ( x ) 3 − q ( x ) 3 = p ( x 3 ) − q ( x 3 ) .
Suppose that p ( x ) − q ( x ) = ( x − 1 ) 5 r ( x ) , where r ( x ) is a polynomial such that x − 1 ∣ r ( x ) . What is the value of s ?
The answer is 9.
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2 solutions
Part i is easy p ( 1 ) 3 − q ( 1 ) 3 = p ( 1 ) − q ( 1 ) . S o p ( 1 ) − q ( 1 ) = 0 .
So ( x − 1 ) divides p ( x ) − q ( x ) .Note s = p ( 1 ) = q ( 1 ) . ( p ( x ) − q ( x ) ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) = p ( x 3 ) − q ( x 3 ) .
So ( x − 1 ) a r ( x ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) = ( x 3 − 1 ) a r ( x 3 ) . Plug x = 1 after cancelling (x-1)^a and we are done.
s 2 = 3 ( a − 1 ) = 3 4 = 8 1 . So s = 9 .
Are you sure there are polynomials that satisfy the conditions of your question?
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Why is there any contradiction???I do not know.Iwill let u know once I construct such a polynomial.
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What you have shown is that if there are polynomials satisfying the conditions, then we must have s = 9 . In order for the answer to not be meaningless, we have to demonstrate that there are such polynomials. Otherwise, any statement about the empty set is vacuously true.
In this case, because the condition is extremely restrictive, I'm concerned about the existence of such polynomials.
s 2 = 8 1 ⇒ s = ± 9 and not only s = 9 .!
Part i is easy p ( 1 ) 3 − q ( 1 ) 3 = p ( 1 ) − q ( 1 ) . S o p ( 1 ) − q ( 1 ) = 0 .
So ( x − 1 ) divides p ( x ) − q ( x ) .Note s = p ( 1 ) = q ( 1 ) . ( p ( x ) − q ( x ) ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) = p ( x 3 ) − q ( x 3 ) .
So ( x − 1 ) a r ( x ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) = ( x 3 − 1 ) a r ( x 3 ) . Plug x = 1 after cancelling (x-1)^a and we are done.
s 2 = 3 ( a − 1 ) = 3 4 = 8 1 . So s = 9 .
@rajdeep brahma , i just edited your solution using latex. You can copy and paste it if you want.
p ( x ) 3 − q ( x ) 3 ( p ( x ) − q ( x ) ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) ( x − 1 ) 5 r ( x ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) ( x − 1 ) 5 r ( x ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) r ( x ) ( p ( x ) 2 + p ( x ) q ( x ) + q ( x ) 2 ) Putting x = 1 ⟹ r ( 1 ) ( p ( 1 ) 2 + p ( 1 ) q ( 1 ) + q ( 1 ) 2 ) r ( 1 ) ( s 2 + s 2 + s 2 ) 3 s 2 s 2 = p ( x 3 ) − q ( x 3 ) = p ( x 3 ) − q ( x 3 ) = ( x 3 − 1 ) 5 r ( x 3 ) = ( x − 1 ) 5 ( x 2 + x + 1 ) 5 r ( x 3 ) = ( x 2 + x + 1 ) 5 r ( x 3 ) = ( 1 2 + 1 + 1 ) 5 r ( 1 3 ) = 3 5 r ( 1 ) = 3 5 = 3 4
s = 9 or s = − 9 (rejected since s < 0)
Note: p ( 1 ) and q ( 1 ) are the sum of the coefficients of p ( x ) and q ( x ) which is easy to prove.