Fun with cube

Algebra Level 2

( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 \large \dfrac{ (a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3 }{ (a-b)^3 + (b-c)^3 + (c-a)^3}

Simplify the expression above.

( a + b + c ) ( a + b + c ) (a+b+c)(a+b+c) ( a + b ) ( c + a ) 3 (a+b)(c+a)^3 ( a + b ) ( b + c ) ( c + a ) (a+b)(b+c)(c+a) a b c abc

Amit Bera by amit bera , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Using : If a + b + c = 0 a+b+c=0 then a 3 + b 3 + c 3 = 3 a b c a^3+b^3+c^3=3abc

We see that so for numerator a 2 − b 2 + b 2 − c 2 + c 2 − a 2 = 0 a^2-b^2+b^2-c^2+c^2-a^2=0 and for denominator we see a − b + b − c + c − a = 0 a-b+b-c+c-a=0 So the equation converts into 3 ( a 2 − b 2 ) ( b 2 − c 2 ) ( c 2 − a 2 ) 3 ( a − b ) ( b − c ) ( c − a ) \frac {3 (a^2-b^2)(b^2-c^2)(c^2-a^2)}{3 (a-b)(b-c)(c-a)} This equation on factorising further simplifies to 3 ( a + b ) ( b + c ) ( c + a ) ( a − b ) ( b − c ) ( c − a ) 3 ( a − b ) ( b − c ) ( c − a ) \frac {3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)}{3 (a-b)(b-c)(c-a)} which on cancelling common terms leaves us with ( a + b ) ( b + c ) ( c + a ) (a+b)(b+c)(c+a)

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution essentially correct, but poorly presented 1. By making a , b , c a, b, c do double duty, it is harder to understand the intention of your first line.
2. The numerator is not a 2 − b 2 + b 2 − c 2 + c 2 − a 2 = 0 a^2-b^2+b^2-c^2+c^2-a^2=0

Suggested presentation:

We know that if x + y + z = 0 x+y+z = 0 , then x 3 + y 3 + z 3 = 3 x y z x^3 + y^3 + z^3 = 3 xyz .

Consider x = a 2 − b 2 , y = b 2 − c 2 , z = c 2 − a 2 x = a^2 - b^2 , y = b^2 - c^2, z = c^2 - a ^2 . This satisfies x + y + z = 0 x + y + z = 0 , hence we can conclude that ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 = 3 ( a 2 − b 2 ) ( b 2 − c 2 ) ( c 2 − a 2 ) (a^2 -b^2)^3 + (b^2-c^2)^3 + (c^2 - a^2) ^3 = 3(a^2-b^2)(b^2-c^2)(c^2-a^2) .

Etc

Amit Bera
Mar 24, 2016

( a ² − b ² ) ³ + ( b ² − c ² ) ³ + ( c ² − a ² ) ³ ( a − b ) ³ + ( b − c ) ³ + ( c − a ) ³ \frac{(a²-b²)³+ (b²-c²)³+(c²-a²)³}{ (a-b)³+(b-c)³+(c-a)³ } = 3 ( a ² − b ² ) ( b ² − c ² ) ( c ² − a ² ) 3 ( a − b ) ( b − c ) ( c − a ) \frac{3(a²-b²) (b²-c²)(c²-a²)}{ 3(a-b)(b-c)(c-a)} =(a+b)(b+c)(c+a)

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Hung Woei Neoh
May 11, 2016

( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 ( a − b ) 3 + ( b − c ) 3 + ( c − b ) 3 = a 6 − 3 a 4 b 2 + 3 a 2 b 4 − b 6 + b 6 − 3 b 4 c 2 + 3 b 2 c 4 − c 6 + c 6 − 3 c 4 a 2 + 3 c 2 a 4 − a 6 a 3 − 3 a 2 b + 3 a b 2 − b 3 + b 3 − 3 b 2 c + 3 b c 2 − c 3 + c 3 − 3 c 2 a + 3 c a 2 − a 3 = − 3 a 4 b 2 + 3 a 2 b 4 − 3 b 4 c 2 + 3 b 2 c 4 − 3 c 4 a 2 + 3 c 2 a 4 − 3 a 2 b + 3 a b 2 − 3 b 2 c + 3 b c 2 − 3 c 2 a + 3 c a 2 = 3 ( − a 4 b 2 + c 2 a 4 + a 2 b 4 − c 4 a 2 − b 4 c 2 + b 2 c 4 ) 3 ( − a 2 b + c a 2 + a b 2 − c 2 a − b 2 c + b c 2 ) = a 4 ( c 2 − b 2 ) − a 2 ( c 4 − b 4 ) + b 2 c 2 ( c 2 − b 2 ) a 2 ( c − b ) − a ( c 2 − b 2 ) + b c ( c − b ) = a 4 ( c 2 − b 2 ) − a 2 ( c 2 + b 2 ) ( c 2 − b 2 ) + b 2 c 2 ( c 2 − b 2 ) a 2 ( c − b ) − a ( c + b ) ( c − b ) + b c ( c − b ) = ( a 4 − a 2 c 2 − a 2 b 2 + b 2 c 2 ) ( c 2 − b 2 ) ( a 2 − a c − a b + b c ) ( c − b ) = ( a 2 ( a 2 − c 2 ) − b 2 ( a 2 − c 2 ) ) ( c 2 − b 2 ) ( a ( a − c ) − b ( a − c ) ) ( c − b ) = ( a 2 − b 2 ) ( a 2 − c 2 ) ( c 2 − b 2 ) ( a − b ) ( a − c ) ( c − b ) = ( a − b ) ( a + b ) ( a − c ) ( a + c ) ( c − b ) ( c + b ) ( a − b ) ( a − c ) ( c − b ) = ( a + b ) ( a + c ) ( c + b ) = ( a + b ) ( b + c ) ( c + a ) \dfrac{(a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3}{(a-b)^3 + (b-c)^3 + (c-b)^3}\\ =\dfrac{\color{#D61F06}{a^6} - 3a^4b^2 + 3a^2b^4 \color{#20A900}{- b^6 + b^6} - 3b^4c^2 + 3b^2c^4 \color{#EC7300}{- c^6 + c^6} - 3c^4a^2 + 3c^2a^4 \color{#D61F06}{- a^6}}{\color{#3D99F6}{a^3} - 3a^2b + 3ab^2 \color{#69047E}{- b^3 + b^3} - 3b^2c + 3bc^2 \color{#624F41}{- c^3 + c^3} - 3c^2a + 3ca^2 \color{#3D99F6}{- a^3}}\\ =\dfrac{\color{#D61F06}{- 3a^4b^2} \color{#20A900}{+3a^2b^4} \color{#EC7300}{- 3b^4c^2 + 3b^2c^4} \color{#20A900}{- 3c^4a^2} \color{#D61F06}{+ 3c^2a^4}}{\color{#3D99F6}{- 3a^2b}\color{#69047E}{ + 3ab^2} \color{#624F41}{- 3b^2c + 3bc^2} \color{#69047E}{- 3c^2a} \color{#3D99F6}{+ 3ca^2}}\\ =\dfrac{3(\color{#D61F06}{-a^4b^2 + c^2a^4} \color{#20A900}{+ a^2b^4 - c^4a^2} \color{#EC7300}{- b^4c^2 + b^2c^4})}{3(\color{#3D99F6}{-a^2b + ca^2}\color{#69047E}{ +ab^2 - c^2a} \color{#624F41}{-b^2c + bc^2})}\\ =\dfrac{\color{#D61F06}{a^4}(c^2-b^2) \color{#20A900}{-a^2}(c^4-b^4) \color{#EC7300}{+b^2c^2}(c^2-b^2)}{\color{#3D99F6}a^2(c-b) \color{#69047E}{-a}(c^2 - b^2) \color{#624F41}{+bc}(c-b)}\\ =\dfrac{a^4\color{cyan}{(c^2-b^2)} -a^2(c^2+b^2)\color{cyan}{(c^2-b^2)} +b^2c^2\color{cyan}{(c^2-b^2)}}{a^2\color{teal}{(c-b)} -a(c+b)\color{teal}{(c - b)} +bc\color{teal}{(c-b)}}\\ =\dfrac{(a^4 -a^2c^2 -a^2b^2 + b^2c^2)\color{cyan}{(c^2-b^2)}}{(a^2 -ac-ab + bc)\color{teal}{(c-b)}}\\ =\dfrac{(a^2\color{magenta}{(a^2 - c^2)} -b^2\color{magenta}{(a^2-c^2)})\color{cyan}{(c^2-b^2)}}{(a\color{#BBBBBB}{(a-c)}-b\color{#BBBBBB}{(a-c)})\color{teal}{(c-b)}}\\ =\dfrac{\color{#CEBB00}{(a^2-b^2)}\color{magenta}{(a^2-c^2)}\color{cyan}{(c^2-b^2)}}{\color{olive}{(a-b)}\color{#BBBBBB}{(a-c)}\color{teal}{(c-b)}}\\ =\dfrac{\color{olive}{(a-b)}\color{#CEBB00}{(a+b)}\color{#BBBBBB}{(a-c)}\color{magenta}{(a+c)}\color{teal}{(c-b)}\color{cyan}{(c+b)}}{\color{olive}{(a-b)}\color{#BBBBBB}{(a-c)}\color{teal}{(c-b)}}\\ =\color{#CEBB00}{(a+b)}\color{magenta}{(a+c)}\color{cyan}{(c+b)}\\ =\boxed{\color{#CEBB00}{(a+b)}\color{cyan}{(b+c)}\color{magenta}{(c+a)}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...