The past, present, and future

Algebra Level 4

Given that E = a 3 − b 3 − c 3 − 3 a b c a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{a^3 -b^3 -c^3 - 3abc}{a^2 + b^2 + c^2 + ab - bc + ac}

and a = 2015 , b = 2016 , c = 2017 a = 2015, b = 2016, c = 2017 . Find E \mathscr{E} .


The answer is -2018.

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Hobart Pao by Hobart Pao , 23, USA

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3 solutions

Chew-Seong Cheong
May 30, 2016

You gotta think positive.

E = a 3 − b 3 − c 3 − 3 a b c a 2 + b 2 + c 2 + a b − b c + a c Let x = a , y = − b , z = − c = x 3 + y 3 + z 3 − 3 x y z x 2 + y 2 + z 2 − x y − y z − z x = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) + 3 x y z − 3 x y z x 2 + y 2 + z 2 − x y − y z − z x = x + y + z = a − b − c = 2015 − 2016 − 2017 = − 2018 \begin{aligned} \mathscr E & = \frac{a^3 -b^3 -c^3 - 3abc}{a^2 + b^2 + c^2 + ab - bc + ac} \quad \quad \small \color{#3D99F6}{\text{Let }x=a, \ y=-b, \ z = -c} \\ & = \frac{x^3+y^3+z^3 - 3xyz}{x^2+y^2+z^2-xy-yz-zx} \\ & = \frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz -3xyz}{x^2+y^2+z^2-xy-yz-zx} \\ & = x+y+z \\ & = a-b-c \\ & = 2015-2016-2017 \\ & = \boxed{-2018} \end{aligned}

14 Helpful 0 Interesting 0 Brilliant 0 Confused

did you use the identity x 3 + y 3 + z 3 − 3 x y z x^3+y^3+z^3 -3xyz expansion

abhishek alva - 5 years ago

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I used x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) + 3 x y z x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz , which is similar.

Chew-Seong Cheong - 5 years ago

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its the same thing

abhishek alva - 5 years ago

Did it the same way sir.+1!

Rishabh Tiwari - 5 years ago

Did the same

Aditya Kumar - 5 years ago

Exactly the same solution. You always have to be on the look out for those types of factorisations.

Sharky Kesa - 5 years ago
Hung Woei Neoh
May 30, 2016

Let b = 2016 = n b=2016=n . Then, c = 2017 = n + 1 , a = 2015 = n − 1 c=2017=n+1,\;a=2015=n-1

Therefore,

E = a 3 − b 3 − c 3 − 3 a b c a 2 + b 2 + c 2 + a b − b c + a c = ( n − 1 ) 3 − n 3 − ( n + 1 ) 3 − 3 ( n − 1 ) ( n ) ( n + 1 ) ( n − 1 ) 2 + n 2 + ( n + 1 ) 2 + ( n − 1 ) ( n ) − ( n ) ( n + 1 ) + ( n − 1 ) ( n + 1 ) = n 3 − 3 n 2 + 3 n − 1 − n 3 − n 3 − 3 n 2 − 3 n − 1 − 3 n 3 + 3 n n 2 − 2 n + 1 + n 2 + n 2 + 2 n + 1 + n 2 − n − n 2 − n + n 2 − 1 = − 4 n 3 − 6 n 2 + 3 n − 2 4 n 2 − 2 n + 1 = − 4 n 3 − 8 n 2 + 2 n 2 + 4 n − n − 2 4 n 2 − 2 n + 1 = 4 n 2 ( − n − 2 ) − 2 n ( − n − 2 ) + 1 ( − n − 2 ) 4 n 2 − 2 n + 1 = ( − n − 2 ) ( 4 n 2 − 2 n + 1 ) 4 n 2 − 2 n + 1 = ( − n − 2 ) = − 2016 − 2 = − 2018 \mathscr{E}\\ =\dfrac{a^3-b^3-c^3-3abc}{a^2+b^2+c^2+ab-bc+ac}\\ =\dfrac{(n-1)^3 - n^3 -(n+1)^3 -3(n-1)(n)(n+1)}{(n-1)^2 + n^2 + (n+1)^2 + (n-1)(n) -(n)(n+1) + (n-1)(n+1)}\\ =\dfrac{\color{#3D99F6}{n^3}\color{#D61F06}{ - 3n^2}\color{#20A900}{+3n}\color{#EC7300}{-1}\color{#3D99F6}{ -n^3 -n^3 }\color{#D61F06}{-3n^2}\color{#20A900}{-3n}\color{#EC7300}{-1}\color{#3D99F6}{-3n^3}\color{#20A900}{+3n}}{\color{#D61F06}{n^2}\color{#20A900}{-2n}\color{#EC7300}{+1}\color{#D61F06}{+n^2+n^2}\color{#20A900}{+2n}\color{#EC7300}{+1}\color{#D61F06}{+n^2}\color{#20A900}{-n}\color{#D61F06}{-n^2}\color{#20A900}{-n}\color{#D61F06}{+n^2}\color{#EC7300}{-1}}\\ =\dfrac{\color{#3D99F6}{-4n^3}\color{#D61F06}{-6n^2}\color{#20A900}{+3n}\color{#EC7300}{-2}}{\color{#D61F06}{4n^2}\color{#20A900}{-2n}\color{#EC7300}{+1}}\\ =\dfrac{\color{#3D99F6}{-4n^3}\color{#D61F06}{-8n^2+2n^2}\color{#20A900}{+4n-n}\color{#EC7300}{-2}}{\color{#D61F06}{4n^2}\color{#20A900}{-2n}\color{#EC7300}{+1}}\\ =\dfrac{4n^2\color{#69047E}{(-n-2)} -2n\color{#69047E}{(-n-2)}+1\color{#69047E}{(-n-2)}}{4n^2-2n+1}\\ =\dfrac{\color{#69047E}{(-n-2)}\color{#624F41}{(4n^2-2n+1)}}{\color{#624F41}{4n^2-2n+1}}\\ =\color{#69047E}{(-n-2)}\\ =-2016-2\\ =\boxed{-2018}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Or use x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) and put x = a , y = − b , z = − c x=a,y=-b,z=-c and question is very trivial since denominator gets cancelled after factorising the numerator as above.

RoYal Abhik - 5 years ago

Another interesting way to look at it.

Hobart Pao - 5 years ago
Hobart Pao
May 30, 2016

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

Then we can add and subtract − 3 a 2 b + 3 a b 2 -3a^2 b+ 3ab^2 .

E = a 3 − b 3 − c 3 − 3 a b c + 3 a 2 b − 3 a b 2 − 3 a 2 b + 3 a b 2 a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{a^3 -b^3 -c^3 - 3abc + 3a^2 b - 3ab^2 - 3a^2 b + 3ab^2 }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b ) 3 − c 3 − 3 a b c + 3 a 2 b − 3 a b 2 a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b)^3 - c^3 -3abc + 3a^2 b - 3ab^2 }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b ) 3 − c 3 + 3 a b ( a − b − c ) a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b)^3 - c^3 + 3ab (a-b-c ) }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b − c ) [ ( a − b ) 2 + c 2 + c ( a − b ) ] + 3 a b ( a − b − c ) a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b-c)[(a-b)^2 + c^2 + c(a-b)] + 3ab (a-b-c ) }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b − c ) [ ( a − b ) 2 + c 2 + c ( a − b ) + 3 a b ] a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b-c)[(a-b)^2 + c^2 + c(a-b) + 3ab] }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b − c ) [ ( a 2 + b 2 − 2 a b + c 2 + a c − b c + 3 a b ] a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b-c)[(a^2 + b^2 - 2ab + c^2 + ac- bc + 3ab] }{a^2 + b^2 + c^2 + ab - bc + ac}

E = ( a − b − c ) [ ( a 2 + b 2 + c 2 + a b + a c − b c ] a 2 + b 2 + c 2 + a b − b c + a c \mathscr{E} = \dfrac{(a-b-c)[(a^2 + b^2 + c^2 + ab + ac- bc ] }{a^2 + b^2 + c^2 + ab - bc + ac}

E = a − b − c \mathscr{E} = a-b-c

Substituting a = 2015 , b = 2016 , c = 2017 a = 2015, b = 2016, c = 2017 , the final answer is 2015 − 2016 − 2017 = − 2018 2015 - 2016 - 2017 = \boxed{-2018}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice question & a nice sol.+1 !

Rishabh Tiwari - 5 years ago

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Thanks for the encouragement!

Hobart Pao - 5 years ago

Too tedious. Could be easily done using x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) . BTW, upvoted.

Rishik Jain - 5 years ago

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