Solving for a + d in Algebraic Equations

[Miike012]

Homework Statement

IF: a/b = c/d

Prove: (a^2c + ac^2) / (b^2d + bd^2) = (a + c)^3/(b+d)^3

The Attempt at a Solution

Let: a/b = c/d = k ; so that a = bk ; c = dk

(a^2c + ac^2) / (b^2d + bd^2) = k^3(b^2d + bd^2) / (b^2d + bd^2)

(a + c)^3/(b+d)^3 = k( b + d) = (b + d)

k^3 =/= k ... what did I do wrong?

 

[Dick]

(kb+kd)^3=k^3*(b+d)^3.

 

[Miike012]

Thank you... I have another one... This does not make sense to me.
If a,b,c,d are proportional, prove
a+d=b+c+((a-b)(a-c)/a)

This is what it said in the book...
a(a-b-c+d) = a^2-ab+ad=a^2 - ab - ac +bc = (a-b)(a-c);
a-b-c+d = +(a-b)(a-c)/a

My question is... how the heck does that prove anything... all the person did is rewrite the problem...??

 

[eumyang]

If a,b,c,d are proportional, prove
a+d=b+c+((a-b)(a-c)/a)

What do you get if you solve
$$a + d = b + c + \frac{(a - b)(a - c)}{a}$$ for the product (a-b)(a-c)?

This is what it said in the book...
a(a-b-c+d) = a^2-ab+ad=a^2 - ab - ac +bc = (a-b)(a-c);

The bolded part is wrong. You are missing a term.

 

[Miike012]

The part you bolded should be... a^2 -ab-ac +ad... now what?

 

[eumyang]

You have
$$a(a - b - c + d) = (a-b)(a-c)$$.
Isolate the a+d inside the parentheses.

 

[Miike012]

a(a+d) - a(c+b) = (a-b)(a-c)

 

[eumyang]

a(a+d) - a(c+b) = (a-b)(a-c)

Well, you could do that, but that wasn't what I had in mind.

Going back to this:
$$a(a - b - c + d) = (a-b)(a-c)$$
how do I isolate the entire expression inside the parentheses?

 

[Miike012]

divide by a =
= a - b - c + d = (a-b)(a-c)/a

is that what you ment?

 

[eumyang]

divide by a =
= a - b - c + d = (a-b)(a-c)/a

is that what you ment?

Yes. Now solve for a + d.