Algebra versus geometry
The sum of the cubes of three consecutive integers = the fourth consecutive integer cubed. After constructing a quadrilateral with the smaller integers opposite each other, rendering the larger integers also opposite one another. Given that the formula for the area of a non-right angled triangle with only the side lengths known is: the square root of s(s - a)(s - b)(s - c); where s = sum of the triangle sides lengths divided by 2. If this area represents a bizarre shaped room in metres, and a builder decides to build 8 such rooms; please find the total area of the 8 quadrilaterals to 3 significant figures.
The answer is 149.
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.
1 solution
Considering that there are four unknowns, but that they are all consecutive integers: p, p+1, p+2 and the product sum p+3; let x = p+1, so as to simplify the equation; then (x - 1)^3 + x^3 + (x + 1)^3 = (x + 2)^3 ==> 3x^3 + 6x^2 = x^3 + 6x^2 + 12x + 8 ==> 2x^3 - 6x^2 - 6x - 8 = 0 = x^3 - 3x^2 - 3x - 4 = 0. So, dividing this expression by x - 4, x^2 + x + 1 is obtained. Therefore (x - 4)(x^2 + x + 1) = 0; so either x^2 + x + 1 = 0; or x - 4 = 0. The first final expression gives an irrational answer for x; therefore x must = 4 = p+1. Therefore p = 3 and the three consecutive integers are 3, 4, 5 and 6. By constructing the quadrilateral with sides 3 and 4 opposite each other, we find that the angle between AB = 3 and BC = 5 is a right angle, so the area of triangle ABC = 7.5. Now, we have to separate off the other triangle ACD by the line AC, which area is then calculated by using the formula in the question and added to that of triangle ABC; from thence to multiply the result by 8, giving 148.9943818: so 149 is the answer to 3 significant figures!