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zeion
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Homework Statement
Prove [tex] x^3 - y^3 = (x - y)(x^2 +xy + y^2) [/tex]
Homework Equations
The Attempt at a Solution
Not sure.. I know I'm not supposed to expand the right side. lol.
Need to use some basic properties?
zeion said:Uhh..
[tex] x^3 - 1 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1 [/tex] ?
zeion said:Okay so one root is x = y
So (x - y) is a root.
zeion said:Um long division?
zeion said:But is there a way to prove this using the properties of addition and multiplication?
zeion said:But is there a way to prove this using the properties of addition and multiplication?
Mentallic said:Uhh, yeah! There is more than one way to prove something, but the easiest would still be to expand the right side
Try it geometrically. First as a warmup, see if you can prove a^{2}-b^{2}=(a-b)(a+b) by drawing a big square of side length a with a small square of side length b inside it (the smaller square should share corners with the larger square).
zeion said:Homework Statement
Prove [tex] x^3 - y^3 = (x - y)(x^2 +xy + y^2) [/tex]
Homework Equations
The Attempt at a Solution
Not sure.. I know I'm not supposed to expand the right side. lol.
Need to use some basic properties?
But the RHS was given. First answer the question asked, then add whatever other information you want. The difficulty here was clearly the statement "I know I'm not supposed to expand the right side. lol."fzero said:What if the RHS wasn't given? Give the guy credit for wanting to learn how to actually learn how to compute something rather than just verify an answer.
HallsofIvy said:But the RHS was given. First answer the question asked, then add whatever other information you want. The difficulty here was clearly the statement "I know I'm not supposed to expand the right side. lol."
This is not true. However, x^{2} - y^{2} = (x - y)(x + y)zeion said:For example in the last question it asked me to prove:
[tex] x^2 + y^2 = (x + y)(x - y) [/tex]
Use = for expressions that are equal, not [itex]\Rightarrow[/itex].zeion said:So I did something like
[tex] x^2 + y^2 \Rightarrow x^2 + xy - xy + y^2 \Rightarrow x(x + y) - y(x + y) \Rightarrow (x + y)(x - y) [/tex]
zeion said:So I was just wondering how to do something similar with this one.
Mentallic said:Uhh, yeah! There is more than one way to prove something, but the easiest would still be to expand the right side
Try it geometrically. First as a warmup, see if you can prove a^{2}-b^{2}=(a-b)(a+b) by drawing a big square of side length a with a small square of side length b inside it (the smaller square should share corners with the larger square).
creillyucla said:Here is a method that doesn't involve manipulating the right hand side, which is more I think of what we're looking for:
Artificially create a factor (x - y) and add canceling terms at the end:
x^3 - y^3 = (x - y)( x^2 ) + ( x - y )( y^2 ) + x^2 * y - y^2 * x
With the final two terms cancelling the terms on the right hand side that don't appear on the left hand side.
Then pull out another ( x - y ) on the final two terms:
x^2 * y - y^2 * x = ( x - y )( x * y )
Put it together and you're done!
The equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) is used to factorize a cubic polynomial into two binomials.
To prove x^3 - y^3 = (x - y)(x^2 +xy + y^2), you can use the distributive property to expand the right side of the equation and then simplify it to match the left side.
The equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) is a special case of the difference of cubes formula, which is commonly used in algebraic manipulation and factoring.
You can use the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) to factorize cubic equations and solve for the roots of the equation.
Yes, the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) can be used in real-life situations such as calculating the volume of a cube or factoring a polynomial in engineering or physics problems.