Find the values of a and b in the given polynomial

[chwala]

Homework Statement

Find the value of $a$ and $b$ in terms of $n$ given;
$(x-n+1)^3-(x-n)^3=3x^2+ax+b$

Relevant Equations

Binomial theorem

My approach;

Let $(n+1)=k$
$(x-n+1)^3-(x-n)^3=3x^2+ax+b$
$(x-k)^3-(x-n)^3=3x^2+ax+b$
$(x^3-3x^2k+3xk^2-k^3)-(x^3-3x^2n+3xn^2-n^3)$
$=-3x^2k+3x^2n+3xk^2-3xn^2-k^3+n^3≡3x^2+ax+b$

$⇒1=n-k$
$⇒3(k^2-n^2)=a$
$⇒n^3-k^3=b$

$3(k+n)(k-n)=a$
$-3(k+n)=a$
$a=-3(n-1+n)$
$a=-3(2n-1)$
$a=3-6n$

$b=n^3-k^3=n^3-(n-1)^3=n^3-n^3+3n^2-3n+1=3n^2-3n+1$
$b=3n^2-3n+1$

which agrees with textbook solution...i would appreciate alternative approach.

 

[anuttarasammyak]

Let x=0 on the both sides
$$b=n^3-(n-1)^3=n^2+n(n-1)+(n-1)^2=3n^2-3n+1$$

Then let x=n on the both sides
$$0=6n^2+(a-3)n$$
$$a=-6n+3$$

 

[Mark44]

Homework Statement:: Find the value of $a$ and $b$ in terms of $n$ given;
$(x-n+1)^3-(x-n)^3=3x^2+ax+b$
Relevant Equations:: Binomial theorem

My approach;

Let $(n+1)=k$
$(x-n+1)^3-(x-n)^3=3x^2+ax+b$
$(x-k)^3-(x-n)^3=3x^2+ax+b$

You have a mistake almost at the very beginning. x - n + 1 = x - (n - 1), so a more reasonable substitution would be to let k = n - 1.
In your work, with your substitution, $(x - k)^3 = (x - (n + 1))^3 = (x - n - 1)^3$. This isn't what you started with.

$(x^3-3x^2k+3xk^2-k^3)-(x^3-3x^2n+3xn^2-n^3)$
$=-3x^2k+3x^2n+3xk^2-3xn^2-k^3+n^3≡3x^2+ax+b$

$⇒1=n-k$
$⇒3(k^2-n^2)=a$
$⇒n^3-k^3=b$

$3(k+n)(k-n)=a$
$-3(k+n)=a$
$a=-3(n-1+n)$
$a=-3(2n-1)$
$a=3-6n$

$b=n^3-k^3=n^3-(n-1)^3=n^3-n^3+3n^2-3n+1=3n^2-3n+1$
$b=3n^2-3n+1$

which agrees with textbook solution...i would appreciate alternative approach.

 

[chwala]

You have a mistake almost at the very beginning. x - n + 1 = x - (n - 1), so a more reasonable substitution would be to let k = n - 1.
In your work, with your substitution, $(x - k)^3 = (x - (n + 1))^3 = (x - n - 1)^3$. This isn't what you started with.

I see your point, what about letting $k=n+1?$ and not $k=(n+1).$ I can see that it's the bracket that causes all the confusion.

 

[Mark44]

which agrees with textbook solution...i would appreciate alternative approach.

I don't see how you arrived at the same solution as the textbook, given the mistake I pointed out. I didn't go through your work beyond what I pointed out, but I suspect that you have made another mistake somewhere.

 

[Delta2]

He says at first line n+1=k but all the rest lines are working like n-1=k, that's why he arrived at the right solution. So practically only his first line is wrong.

 

[Prof B]

Let x=0 on the both sides
...

f(x) is a polynomial. Its constant term is f(0).

Then let x=n on the both sides
...

Alternatively, the coefficient of x is f'(0). That's not pre-calc, but he did ask for an alternative solution.

 

[chwala]

f(x) is a polynomial. Its constant term is f(0).Alternatively, the coefficient of x is f'(0). That's not pre-calc, but he did ask for an alternative solution.

@Prof B I do not think you have indicated any improvement on the problem nor alternative solution. Can you show your working by using $f'(0)$ as you have pointed out?

 

[Fred Wright]

Try expanding the cubic term with binomial theorem like this
$$
((x-n)+1)^3=(x-n)^3 + 3(x-n)^2 + 3(x-n) +1
$$
and observe that the $(x-n)^3$ gets subtracted off on the r.h.s. In other words; think ahead. I think that was the point of this exercise.

 

[Prof B]

@Prof B I do not think you have indicated any improvement on the problem nor alternative solution. Can you show your working by using $f'(0)$ as you have pointed out?

f'(0) is the derivative of f(x) evaluated at 0. It will make sense when you take calculus.

 

[Delta2]

f'(0) is the derivative of f(x) evaluated at 0. It will make sense when you take calculus.

He knows calculus, he is 43 years old . You can check additional info for each member, either by clicking at the avatar or hovering the mouse over it. Just saying cause you are new at these forums.
And welcome to PF from me, first indication is that you are quite knowledgeable about math!

 

[Prof B]

I'm not sure how I was supposed to infer that. Pythagoras was a great mathematician. He lived to the age of 75 and did not know calculus.

 

[Delta2]

I'm not sure how I was supposed to infer that. Pythagoras was a great mathematician. He lived to the age of 75 and did not know calculus.

Yes ok I understand that. You don't know @chwala, he is quite sometime in these forums and has post many problems over a wide range of math topics, including calculus.

 

[Mark44]

Homework Statement:: Find the value of $a$ and $b$ in terms of $n$ given;
$(x-n+1)^3-(x-n)^3=3x^2+ax+b$
Relevant Equations:: Binomial theorem

My approach;

Let $(n+1)=k$

The equation above is wrong. You want to replace x - n + 1 by x - k, so k = n - 1.

$(x-n+1)^3-(x-n)^3=3x^2+ax+b$
$(x-k)^3-(x-n)^3=3x^2+ax+b$
$(x^3-3x^2k+3xk^2-k^3)-(x^3-3x^2n+3xn^2-n^3)$
$=-3x^2k+3x^2n+3xk^2-3xn^2-k^3+n^3≡3x^2+ax+b$

$⇒1=n-k$

This isn't equivalent to n + 1 = k, which was the substitution you made several lines up.

$⇒3(k^2-n^2)=a$
$⇒n^3-k^3=b$

In addition to the comments above, the work above seems incomplete. The 2nd and 3rd equations above come from equating the x terms and constant terms on both sides of the equation above the one that says 1 = n - k. The equation that seems to be missing is the one that equates the $x^2$ terms on both sides. Once I had these three equations, then I would solve for a and b.

$3(k+n)(k-n)=a$
$-3(k+n)=a$

Here's another mistake that, fortunately for you, cancels out your earlier mistake. According to your substitution (which is incorrect), k = n + 1. This means, with that substitution, that the 2nd equation above should be $3(k + n)(n + 1 - n) = a$, or $3(k + n)(1)$.

$a=-3(n-1+n)$
$a=-3(2n-1)$
$a=3-6n$

$b=n^3-k^3=n^3-(n-1)^3=n^3-n^3+3n^2-3n+1=3n^2-3n+1$
$b=3n^2-3n+1$

which agrees with textbook solution...i would appreciate alternative approach.

 

[chwala]

@Mark44 hi, let me look at it again...Will get back on this.

 

[chwala]

Ok here we have it;...pretty easy

$x-n+1)^3-(x-n)^3=3x^2+ax+b$

Re-arranging;

$(x+1-n)^3-(x-n)^3=3x^2+ax+b$

Using Combinations-

$[^3C_0x^3+3C_1x^2 (1-n)+3C_2x (1-n)^2 +3C_3x^0 (1-n)^3]-[3C_0x^3+3C_1x^2(-n)+3C_2x(-n)^2+3C_3x^0 (-n)^3]$

$=[x^3+3x^2(1-n)+3x(1-n)^2+(1-n)^3]-[x^3-3x^2n+3xn^2-n^3]$

$=[x^3+3x^2-3x^2n+3x-6xn+3xn^2-n^3+3n^2-3n+1]-[x^3-3x^2n+3xn^2-n^3]$

$=3x^2+3x-6xn+3n^2-3n+1$

$=3x^2+(3-6n)x+3n^2-3n+1$

therefore,

$a=3-6n$

$b=3n^2-3n+1$