MHB Yuri's questions at Yahoo Answers involving algebra

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Yuri's algebra questions involve two main problems related to liquid flow and the factorization of cubes. For the first problem, the values of p and q are determined to be 300 and 0, respectively, and the coefficients a, b, and c are found to be 2, 1, and -4 through the identity 300-A(t)=B(t). The second problem requires factorizing the difference of cubes of two consecutive positive odd numbers, leading to the conclusion that this difference is always odd and thus never divisible by 4. The discussion includes detailed calculations and explanations for each step. Overall, the thread provides a comprehensive breakdown of the algebraic concepts involved.
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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.
 
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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.
 
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