Oh no cubes!

Algebra Level 2

( 28 ) 3 + ( − 15 ) 3 + ( − 13 ) 3 (28)^{ 3 }+(-15)^{ 3 }+(-13)^{ 3 }


The answer is 16380.

A Former Brilliant Member by A Former Brilliant Member

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.

7 solutions

Ananya Maheshwari
Oct 18, 2016

Relevant wiki: Algebraic Identities

If a+b+c=0, then a³+b³+c³=3abc
Here, a=28; b=-15; c=-13
So a+b+c=28-15-13=0
So a³+b³+c³=3x28x(-15)x(-13)=16380


94 Helpful 0 Interesting 0 Brilliant 0 Confused

@Ellea Demko If a+b+c=0, then a³+b³+c³=3abc is a valid algebraic identity. You can check it in the link given below. (Point 32) http://www.mathformula.in/mobile/algebra.html

Ananya Maheshwari - 4 years, 3 months ago

well you man

sunshine 张 - 4 years, 7 months ago

you are all wrong....cube root both side...and zero side.....equal 0....Nightmare Before Christmas....Zero

Robbie Guerrero - 4 years, 4 months ago

3abc is your first mistake and so all following steps are invalid

Ellea Demko - 4 years, 4 months ago
Miae Kim
Oct 18, 2016

2 8 3 = ( 15 + 13 ) 3 28^3 = (15+13)^3 = 1 5 3 + 3 ∗ 13 ∗ 15 ( 13 + 15 ) + 1 3 3 = 15^3 + 3*13*15(13+15) + 13^3

Therefore the original formula become 3 ∗ 13 ∗ 15 ( 13 + 15 ) = 16380 3*13*15(13+15) = 16380

60 Helpful 0 Interesting 0 Brilliant 0 Confused

This is the way I did it, and it's right. Trust me.

Jeff Smith - 4 years, 4 months ago

There's literally no 3* anything. Multiplying 3 is not part of the equation at any point in time

Ellea Demko - 4 years, 4 months ago

wrong---zero

Robbie Guerrero - 4 years, 4 months ago
Anshu Garg
Oct 31, 2016

28-15-13=0 a+b+c=0 then a^3+b^3+c^3=3abc answer is 16380

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Exponents come before subtraction and there isn't even any subtraction in this problem

Ellea Demko - 4 years, 4 months ago

Log in to reply

its an identity

anshu garg - 4 years, 4 months ago
Amit Kumar
Oct 23, 2016

a=-15, b=13; Expand {a+b}^3 ==> {a}^3+{b}^3+3 a b (a+b) [by binomial]; ==>{a+b}^3 - 3 a b (a+b) = {a}^3+{b}^3 --(1); subtracting (1) from 28^3 on both sides; ==> 28^3-{a+b}^3-3 a b (a+b)=28^3-{a}^3-{b}^3; ==> -3 a b (a+b) = Required result; ==> -3 -15 -13*{-15-13}; ==>16380

7 Helpful 0 Interesting 0 Brilliant 0 Confused

And, yet, it still seems easier to term the entire thing into powers of 2--messier, but less rote in its focus.

Joshua Nesseth - 4 years, 7 months ago

wrong, zero....F -

Robbie Guerrero - 4 years, 4 months ago

Log in to reply

Mr. President, it's nice to see you've graduated from Twitter to more...varied media.

Joshua Nesseth - 4 years, 4 months ago
Yash Ghaghada
Feb 27, 2017

Try to see the simple relations that you can use to minimize calculations, as here The bigger is the sum of small two, so you can cancel the cubes if 15 and 13 By writing 28=15+13

0 Helpful 0 Interesting 0 Brilliant 0 Confused

It doesn't say anything about not using a calculator. ;)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Albert Kirsch
Nov 29, 2016

Less elegant, but one could simply plug it into a calculator.....

0 Helpful 0 Interesting 0 Brilliant 0 Confused

That's how I worked it out!

Paul Townsend - 4 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...