Yuri's questions at Yahoo Answers involving algebra

[MarkFL]

Here are the questions:

Maths problem please help me?

1.) A liquid flows from container A to B. The volumes, in cm^3 of the liquid in the containers A and B, t seconds after the start of the experiment are

A(t)=-t^3+at^2+p
B(t)=bt(t-2)^2+2t^2+ct+q

At the start of the experiment container A has 300cm^3 and B is empty. Find the values of p and q.

Explain why 300-A(t)=B(t) is in identity for all non-negative values of t up to the instant when container A is empty.( I don't know what the question means)

Use the identity above to find value of a, b and c

2. Factorize (2n+1)^3-(2n-1)^3 completely and show that the difference between the cubes of two consecutive positive odd numbers can never be divisible by 4.

I have posted a link to this topic so the OP can see my work.

 

[MarkFL]

Hello Yuri,

1.) We are given the functions:

$$A(t)=-t^3+at^2+p$$

$$B(t)=bt(t-2)^2+2t^2+ct+q$$

Now, we are also given:

$$A(0)=300=-(0)^3+a(0)^2+p=p\,\therefore\,p=300$$

$$B(0)=0=b(0)(0-2)^2+2(0)^2+c(0)+q=q\,\therefore\,q=0$$

Thus, we have:

$$A(t)=-t^3+at^2+300$$

$$B(t)=bt(t-2)^2+2t^2+ct$$

Because the amount of liquid present in A plus the amount present in B has to remain constant (assuming so spills, leaks, evaporation, etc.), we know:

$$A(t)+B(t)=300$$

Which we can arrange as:

$$300-A(t)=B(t)$$

Using the definitions of the functions, we may write:

$$300-\left(-t^3+at^2+300 \right)=bt(t-2)^2+2t^2+ct$$

Distributing, expanding and collecting like terms, we have:

$$t^3-at^2=bt^3+(2-4b)t^2+(4b+c)t$$

Equating coefficients, we find:

$$b=1$$

$$4b-2=a\,\therefore\,a=2$$

$$4b+c=0\,\therefore\,c=-4$$

2.) Using the difference of cubes formula, we may write:

$$(2n+1)^3-(2n-1)^3=((2n+1)-(2n-1))\left((2n+1)^2+(2n+1)(2n-1)+(2n-1)^2 \right)=2\left(12n^2+1 \right)$$

The second factor was obtained by noting:

$$(2n+1)^2=4n^2+4n+1$$

$$(2n+1)(2n-1)=4n^2-1$$

$$(2n-1)^2=4n^2-4n+1$$

Adding, we find:

$$(2n+1)^2+(2n+1)(2n-1)+(2n-1)^2=12n^2+1$$

Since $$12n^2+1$$ is odd for any $n\in\mathbb{N}$, we find then that the difference between the cubes of two consecutive positive odd numbers can never be divisible by 4.

If we let $$m=6n^2$$ then $$12n^2+1=2m+1$$, which is obviously odd.