- #1

- 18

- 0

I also would like to know how do you obtain ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)## and ##a^3 + b^3 = (a + b)(a^2 - ab + b^2)##?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- B
- Thread starter gede
- Start date

- #1

- 18

- 0

I also would like to know how do you obtain ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)## and ##a^3 + b^3 = (a + b)(a^2 - ab + b^2)##?

- #2

SteamKing

Staff Emeritus

Science Advisor

Homework Helper

- 12,798

- 1,670

Factoring a

I also would like to know how do you obtain ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)## and ##a^3 + b^3 = (a + b)(a^2 - ab + b^2)##?

Factoring a

For the factoring of the cubics (or polynomials in general), you can always use polynomial long division.

- #3

Mark44

Mentor

- 35,287

- 7,140

And if you're limited to factors with real coefficients, it's not factorable at all.SteamKing said:Factoring a^{4}+ b^{4}is tricky.

- #4

- 299

- 20

And if you're limited to factors with real coefficients, it's not factorable at all.

Actually it is. It factors as [itex](a^2 + \sqrt{2}ab + b^2)(a^2 - \sqrt{2}ab + b^2)[/itex]

- #5

member 587159

Substituting a^2 = x and b^2 = y

=> x^2 - y^2 = ...

2) a^4 + b^4 = a^4 + 2a^2*b^2 + b^4 - 2a^2*b^2 = (a^2 + b^2)^2 - 2a^2*b^2

Now use a^2 - b^2 = (a-b)(a+b) to find the answer

3) a^3 - b^3 = (a-b)(a^2 + ab + b^2)

=> (a^3 - b^3 )/(a-b) = (a^2 + ab + b^2)

Use euclidean division or Horner (let a^3 be the variable, let b^3 be the constant)

- #6

- 12

- 0

1) The first factor, which is a binomial, is always the cube root of the two terms.

(The blank space represents the missing terms and its operation/ sign).

a

2) The second factor, which is a trinomial, always constitutes three terms as follows.

a. The first term is square of the first term in the binomial: (a - b)(a

b. The middle term is reverse sign of the product of the two term: (a - b)(a

c. The last term is the square of the second term in the binomial: (a - b)(a

And that completes the factoring of (a

- #7

member 587159

^{3}- b^{3}) and any expression alike of this. (Only works for this kind of binomial).

1) The first factor, which is a binomial, is always the cube root of the two terms.

(The blank space represents the missing terms and its operation/ sign).

a^{3}- b^{3}= (a - b)( _ _ _ )

2) The second factor, which is a trinomial, always constitutes three terms as follows.

a. The first term is square of the first term in the binomial: (a - b)(a^{2}_ _ )

b. The middle term is reverse sign of the product of the two term: (a - b)(a^{2}+ ab _)

c. The last term is the square of the second term in the binomial: (a - b)(a^{2}+ ab + b^{2})

And that completes the factoring of (a^{3}- b^{3})

This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.

- #8

- 12

- 0

This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.

Nonetheless, it is a basic format on how to factor such kind of expressions. It's not a point of remembering a trick, it is one way of knowing a proof how such expression can be branched down into its respective factors.

- #9

Mark44

Mentor

- 35,287

- 7,140

It's a huge time saver to remember a few "tricks", such as the Quadratic Formula and how to factor the difference of squares and the sum or difference of cubes.This is true, although the point of mathematics is not to remember some tricks.

You don't need calculus to derive ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)##. Also, since the OP is studying algebra, it's reasonable to assume that he hasn't been exposed to the techniques of calculus.Math_QED said:With some experience, it takes 30 seconds to proof this formula using general calculus.

- #10

- 22,129

- 3,297

This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.

How would you prove this using calculus?

- #11

- 18

- 0

Please tell me the method of factoring of ##a^n + b^n## and ##a^n - b^n##?

- #12

member 587159

Please tell me the method of factoring of ##a^n + b^n## and ##a^n - b^n##?

1) Define a function F: R -> R: a -> a^n + b^n => F(a) = a^n + b^n

b and n are natural numbers, n is odd

We see: F(-b) = -b^n + b^n = 0 => a + b is a divisor of F(a) (fundamental theorem of algebra)

That's one factor, you can find what's left after the division using horner's rule or using euclidean division. Then you can try to find those factors.

Same thing when you have Z(a) = a^n -b^n

b and n natural numbers, n odd

2) Define a function G: R -> R: a -> a^n - b^n => G(a) = a^n - b^n

b and n are natural numbers, n is not odd

Use a^2 - b^2 = (a-b)(a+b) to find factors.

3) Define a function H: R -> R: a -> a^n + b^n => H(a) = a^n + b^n

b and n are natural numbers, n is not odd

This is the hardest one. You have to manipulate the expressions. For example,

a^4 +1 = a^4 + 1 + 2a^2 - 2a^2 = (a^2 +1)^2 - 2a^2 = (a^2 + 1 + SQRT(2)a)(a^2 + 1 - SQRT(2)a)

- #13

Mark44

Mentor

- 35,287

- 7,140

None of these techniques use calculus, which is what @micromass asked about. For ##a^4 + 1##, if by factoring, one means splitting the polynomial into1) Define a function F: R -> R: a -> a^n + b^n => F(a) = a^n + b^n

b and n are natural numbers, n is odd

We see: F(-b) = -b^n + b^n = 0 => a + b is a divisor of F(a) (fundamental theorem of algebra)

That's one factor, you can find what's left after the division using horner's rule or using euclidean division. Then you can try to find those factors.

Same thing when you have Z(a) = a^n -b^n

b and n natural numbers, n odd

2) Define a function G: R -> R: a -> a^n - b^n => G(a) = a^n - b^n

b and n are natural numbers, n is not odd

Use a^2 - b^2 = (a-b)(a+b) to find factors.

3) Define a function H: R -> R: a -> a^n + b^n => H(a) = a^n + b^n

b and n are natural numbers, n is not odd

This is the hardest one. You have to manipulate the expressions. For example,

a^4 +1 = a^4 + 1 + 2a^2 - 2a^2 = (a^2 +1)^2 - 2a^2 = (a^2 + 1 + SQRT(2)a)(a^2 + 1 - SQRT(2)a)

That was what I was thinking when I said that ##a^4 + b^4## wasn't factorable. To clarify my statement, ##a^4 + b^4## isn't factorable into linear factors with real coefficients.

Share: