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If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$

Vlad
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Ajay Sharma
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2 Answers2

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HINT: Substitute $\,f(x) = ax^3 + bx^2 + cx + d$ into your equation and collect coefficients of the different powers of $x$. Solve obtained system of linear equations, thus getting an explicit expression for $f$. Substitute $\,x = 5\,$ into obtained equation and get the answer.

How do we know that $f$ is the third-order polynomial?

On the left we have expression involving $\;f\left(x^3\right),\,$ and on the right $\;f^2\left(x\right).\,$ Assuming $\,f\,$ is a polynomial of degree $\,n\in\mathbb R,\,$ we conclude that $\,3\left(n-1\right) = 2n.\,$ Therefore we conclude that polynomial $\,f\,$ is of order $3$, i.e. $\;f(x) = ax^3 + bx^2 + cx + d.\,$

EDIT: Upon request of @AjaySharma I provide explicit solution: $$ f(x) = a x^3 + b x^2 + c x + d \implies \begin{cases} f\left(x^3\right) = a x^9 + b x^6 + c x^3 + d \\ f\left(x^3 - 2\right) = a \left(x^3-2\right)^3 + b \left(x^3-2\right)^2 + c \left(x^3-2\right) + d \end{cases} $$ Thus the difference $$ \begin{aligned} f\left(x^3\right)-f\left(x^3-2\right) & = a \left(\left(x^3\right)^3 - \left(x^3-2\right)^3 \right) + b \left( \left(x^3\right)^2 - \left(x^3-2\right)^2 \right) + c \left( \left(x^3\right) - \left(x^3-2\right) \right) \\ & = 2a \left(\left(x^3\right)^2 + \left(x^3\right) \left(x^3-2\right)+ \left(x^3-2\right)^2 \right) + 2b \left( 2x^3-2 \right) + 2c \\ & = 2a \left({x^6} + {x^6} - {2x^3} + {x^6} - {4x^3} + 4 \right) + 4b \left( x^3-1 \right) + 2c \\ & = 6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c \end{aligned} $$

On the other hand, $$ \begin{aligned} f^2\left(x\right) & = \left(a x^3 + b x^2 + c x + d \right)^2 = \left(a x^3 + b x^2 + c x + d \right)\cdot \left(a x^3 + b x^2 + c x + d \right) = \\ & = a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2 \end{aligned} $$

Therefore $\qquad f\left(x^3\right)-f\left(x^3-2\right) = f^2\left(x\right)+12 \implies $ $$ \begin{aligned} \implies& \ 6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c = \\ =&\ a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2 + 12 \end{aligned} $$ Let us gather coefficients by powers of $x$: $$ \begin{aligned} &x^6:& a^2 &= 6 a &\implies& & a &= 6 \\ &x^5:& 2ab &= 0 &\implies& & b &= 0 \\ &x^4:& b^2+2ac &= 0 &\implies& & c &= 0 \\ &x^3:& 2ad + 2bc &= 4 b - 12 a &\implies& & d &= -6 \\ &x^2:& c^2 + 2bd &= 0 \\ &x^1:& 2cd &= 0 \\ &x^0:& d^2 + 12 &= 8a - 4b + 2c \end{aligned} $$

Thus we get $$ \bbox[5pt, border:2.5pt solid #FF0000]{f\left(x\right) = 6x^3-6} $$ and so $\,f\left(5\right) = 744 .\,$

Vlad
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  • Thats false; we only now $deg(f(x^3)-f(x^3-2))≤3deg(f)$. – Redundant Aunt Aug 23 '15 at 11:45
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    LHS has order $3(n-1)$ while RHS has order $2n$, hence we find $n=3$. This is because the highest order term on LHS gets cancelled when you do $f(x^3)-f(x^3-2)$. – Macavity Aug 23 '15 at 11:48
  • what if 3(n-1) also get cancelled @Macavity – Ajay Sharma Aug 23 '15 at 11:54
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    @AjaySharma then the coefficient $,a,$ in front of $,x^3,$ will be equal to zero. – Vlad Aug 23 '15 at 11:56
  • @AjaySharma that may rarely happen in which case you get a lower order polynomial, but the highest term will always get cancelled. Which is all that you need here. – Macavity Aug 23 '15 at 11:57
  • With this way it will require lot of calculation can any one tell me other way to solve this – Ajay Sharma Aug 23 '15 at 14:39
  • @AjaySharma the calculations are not bad at all. In fact, I have typed up the full solution and the answer for your convenience. – Vlad Aug 23 '15 at 15:53
  • Is it obvious that coefficient of $x^{3(n-1)}$ will not cancel? It is when you write $f$ as $\sum_k a_kx^k$ and expand $f(x^3)-f(x^3-2)$ (the coefficient of $x^{3(n-1)}$ is $2na_n$) but can we somehow see it directly? – Ennar Aug 23 '15 at 18:08
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Given

$$ f(x^3) - f(x^3-2) = f(x)^2 + 12. \tag 1 $$

Let $f(x) = P_n(x)$, where $P_n(x)$ is a polynomial of degree $n$. LHS is of degree $3(n-1)$ and RHS is of degree $2n$. Whence

$$ \bbox[16px,border:2px solid #800000] { f(x) = P_3(x) }. \tag 2 $$

We can differentiate $(1)$ and we obtain

$$ \begin{eqnarray} 3 x^2 \Big[ f'(x^3) - f'(x^3-2) \Big] &=& 2 f(x) f'(x)\\\\ 3 \Big[ 2 x + 3 x^4 \Big] \Big[ f'(x^3) - f'(x^3-2) \Big] &=& 2 f(x) f''(x) + 2f'(x)^2\\\\ 3 \Big[ 2 + 18 x^3 + 9 x^6 \Big] \Big[ f'(x^3) - f'(x^3-2) \Big] &=& 2 f(x) f'''(x) + 6 f'(x) f''(x) \end{eqnarray} $$

Put in $x=0$ so we get

$$ \begin{eqnarray} f(0) f'(0) &=& 0\\\\ f(0) f''(0) + f'(0)^2 &=& 0\\\\ f(0) f'''(0) + 3 f'(0) f''(0) &=& 3 \Big[ f'(0) - f'(-2) \Big] \end{eqnarray} $$

However, $f'(0) \ne 0$ gives a contradiction, as $f(0)=0$ and then $f(0) f''(0) + f'(0)^2=0$. So

$$ \bbox[16px,border:2px solid #800000] { f'(0) = 0 } \tag 3 $$

If $f(0)=0$ we get $f'(0) = f'(-2) = 0$, which implies that $f(x) = c$ - which is to be excluded. So

$$ \bbox[16px,border:2px solid #800000] { f(0) \ne 0 } \tag 4 $$

As $f(0) f''(0) + f'(0)^2 = 0$. So

$$ \bbox[16px,border:2px solid #800000] { f''(0) = 0 } \tag 5 $$

And we also get

$$ f'''(0) = - 3\frac{ f'(-2) }{ f(0) } \tag 6 $$

From $(2)$ $(3)$ $(4)$ and $(5)$ follows that

$$ f(x) = p x^3 + q \tag 7 $$

From $(7)$ follows

$$ 6 p = - \frac{36p}{q}, \quad \Rightarrow \quad q = -6. \tag 8 $$

Put in $x=0$ in $(1)$ and we get

$$ f(0) - f(-2) = f(0)^2 + 12. $$

Whence

$$ - 6 + 8 p + 6 = 6^2 + 12, \quad \Rightarrow \quad p = 6. \tag 9 $$

So we obtain

$$ \bbox[16px,border:2px solid #800000] { f(x) = 6 x^3 - 6 } \quad \Rightarrow \quad \bbox[16px,border:2px solid #800000] {f(5) = 6 \times 5^3 - 6 = 744 } $$

johannesvalks
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