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Doubling the cube is the construction, using only a straightedge and compass, of the edge of a cube that has twice the volume of a cube with a given edge. This is impossible because the cube root of 2, though algebraic, cannot be computed from integers by addition, subtraction, multiplication, division, and taking square roots.
Squares and Cubes. Brahmagupta then goes on to give the sum of the squares and cubes of the first n integers. 12.20. The sum of the squares is that [sum] multiplied by twice the [number of] step[s] increased by one [and] divided by three. The sum of the cubes is the square of that [sum] Piles of these with identical balls [can also be computed].
Doubling the cube, also known as the Delian problem, is an ancient [a] [1] : 9 geometric problem. Given the edge of a cube, the problem requires the construction of the edge of a second cube whose volume is double that of the first. As with the related problems of squaring the circle and trisecting the angle, doubling the cube is now known to ...
The other roots of the equation are obtained either by changing of cube root or, equivalently, by multiplying the cube root by a primitive cube root of unity, that is . This formula for the roots is always correct except when p = q = 0 , with the proviso that if p = 0 , the square root is chosen so that C ≠ 0 .
In geometry, a tesseract or 4-cube is a four-dimensional hypercube, analogous to a two-dimensional square and a three-dimensional cube. Just as the perimeter of the square consists of four edges and the surface of the cube consists of six square faces , the hypersurface of the tesseract consists of eight cubical cells , meeting at right angles .
Vieta's formulas can equivalently be written as. for k = 1, 2, ..., n (the indices ik are sorted in increasing order to ensure each product of k roots is used exactly once). The left-hand sides of Vieta's formulas are the elementary symmetric polynomials of the roots. Vieta's system (*) can be solved by Newton's method through an explicit ...
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