We can figure this out algebraically.\u00a0You may be inspired to see that:<\/p>\n
(x + y)3<\/sup> = (x + y)(x + y)2<\/sup> = (x + y)(x2<\/sup> + 2xy + y2<\/sup>)<\/i><\/p>\n And then we distribute:<\/p>\n (x + y)(x2<\/sup> + 2xy + y2<\/sup>) = <\/i><\/p>\n \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x(x2<\/sup> + 2xy + y2<\/sup>) +\u00a0 y(x2<\/sup> + 2xy + y2<\/sup>) = <\/i><\/p>\n \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x3<\/sup> +2x2<\/sup>y + xy2<\/sup> + x2<\/sup>y + 2xy2<\/sup> + y3<\/sup><\/i><\/p>\n And then you group the common terms to discover the rule for the cube of a binomial.<\/p>\n (x + y)3<\/sup> = x3<\/sup> + 3x2<\/sup>y +\u00a0 3xy2<\/sup> + y3<\/sup><\/i><\/strong><\/p>\n Let me just say that you have a rare appreciation for algebra if that equation by itself stirs you. \u00a0And I would not be shocked to find that you skipped over the algebra that led to it.\u00a0 \u00a0But once again, it\u2019s better with pictures.<\/p>\n This time, we build a cube with each edge x + y<\/i> units long:<\/p>\n Then, we take that cube apart and see what we have:<\/p>\n Let\u2019s take inventory.\u00a0 The big cube we started with has been broken up into the following collection:<\/p>\n But the volume of the collection has to add up to the volume of the big cube we started with.<\/p>\n That means:\u00a0 (x + y)3<\/sup> = x3<\/sup> + 3x2<\/sup>y +\u00a0 3xy2<\/sup> + y3<\/sup><\/i> !\u00a0 <\/strong><\/p>\n The 3-d picture makes this equation\u00a0worthy of an exclamation point. The picture helps you to “see” the equation.\u00a0 That can also work the other way around.\u00a0 Let’s consider a 4-dimensional “hypercube” with edges that are (x + y).\u00a0 I can’t picture\u00a0those.\u00a0 But I can still get some insight about them from the algebraic approach.\u00a0 I’ll skip the steps this time and go right to the result:<\/p>\n (x + y)4<\/sup> = x4<\/sup> + 4x3<\/sup>y + 6x2<\/sup>y2<\/sup> + 4xy3<\/sup> + y4<\/sup><\/strong><\/em><\/p>\n Even though I can’t picture a 4-dimensional cube, I think I know what happens when you break it into pieces: you get 1 big hypercube, 1 smaller hyper cube and a variety of hyper-rectangular solids.\u00a0 You can predict the\u00a0dimensions and the number of each type from the equation above.\u00a0\u00a0So I guess this time, it’s better with equations.<\/p>\n (The graphics in this post were created in Google Sketch-up by my daughter.)<\/p>\n","protected":false},"excerpt":{"rendered":" Well, I hope that the pictures in my last post changed the way you look at the square of a binomial.\u00a0 And I am sure that I know what you are thinking now: “What about binomials, cubed?” We can figure … Continue reading ![\"cube2\"](\"http:\/\/advancedmathyoungstudents.com\/blog\/wp-content\/uploads\/2014\/03\/cube2.jpg\")
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