Help with algebra?

Question

There are three rational expressions:

ax^2/[(x - a)(x - b)(x - c)] ------------------------- (1)
bx/[(x - b)(x - c)] ---------------------------- (2)
c/(x - c) -------------------------- (3)

Prove that:
(1) + (2) + (3) + 1 = x^3/[(x - a)(x - b)(x - c)]

I added the expressions. I get x^3 + 2ax in the numerator. I want to know the mistake I made in adding.

The (1) refers to the first expression and the last '1' is the natural number 1.

Please add everything. I made an addition mistake. I got everything right till the ax^2 + bx^2... step. Things went downhill from there.

The dots appear when too many characters are typed without a space. The other characters are converted into ...

Answer 1

ax^2/[(x - a)(x - b)(x - c)] + bx/[(x - b)(x - c)] + c/(x - c) + [(x - a)(x - b)(x - c)]/[(x - a)(x - b)(x - c)]
note that [(x - a)(x - b)(x - c)]/[(x - a)(x - b)(x - c)] is equal to 1.

then for each you multiply by the denominators and nominators that each expression needs in order to have the same denominator that is [(x - a)(x - b)(x - c)]

therefore, ax^2/[(x - a)(x - b)(x - c)] + bx(x-a)/[(x - a)(x - b)(x - c)] + c(x-a)(x-b)/[(x - a)(x - b)(x - c)] + [(x - a)(x - b)(x - c)]/[(x - a)(x - b)(x - c)] = ax^2+bx^2-abx+cx^2+-bcx-acx+abc+x^3-ax^2-bx^2+abx+acx-abc/[(x - a)(x - b)(x - c)] and doing all the sums and subtractions, you should get x^3/[(x - a)(x - b)(x - c)]

hope you understand it

Answer 2

Use substitution to replace the first "1" with the first rational expression, replace "2" with the second rational expression, replace "3" with the third rational expression and replace the last "1" with the first rational expression. Then add the four fractions together by creating a common denominator of (x-a)(x-b)(x-c). The algrebra isn't "nice" but it isn't bad either. All of the terms in the numerator add/subtract away except for x^3. Good luck!

Answer 3

its simple
take the lcm of ist three eqns.. it comes out to be

ax^2+bx^2+cx^2-abx-acx-bcx+abc/(x-a)(x-b)(x-c).
this is result of (1)+(2)+(3).
Now add this to 1
when you expend (x-a)(x-b)(x-c), it comes out to be
x^3-ax^2-bx^2-cx^2+abx_acx_bcx-abc.
Now you get the required answer by adding both terms with (x-a)(x-b)(x-c) in denominatior.

Answer 4

1.

ax^2/(x-a)(x-b)(x-c) + bx(x-a)/(x-a)(x-b)(x-c) + c(x-a)(x-b)/(x-a)(x-b)(x-c) + (x-a)(x-b)(x-c)/(x-a)(x-b)(x-c)

2. Common Denominator

ax^2+bx(x-a)+c(x-a)(x-b)+(x-a)(x-b)(x-c)
------------------------------------------------------
(x-a)(x-b)(x-c)

3. Expanding Numerator
ax^2+bx^2-abx+cx^2-xbc-axc+abc+x^3-x^2c-bx^2+bxc-ax^2+axc+abx-abc

4. Crossing Out (+ with -)
ax^2-abx-acx+abc-ax^2+axc+abx-abc= 0
bx^2+cx^2-xbc+x^3-x^2c-bx^2+bxc=x^3


5. Solution
x^3
---------------------------
[(x - a)(x - b)(x - c)]