Elements of Algebra

CHAP. XI. OF ALGEBRA. 259 If the less be represented by $x$, the greater will evidently be $x + 720$; and, by the question, $x \sqrt{(x + 720)} = 20736 = 8 \cdot 8 \cdot 4 \cdot 81$. Squaring both sides, we have $x^2(x + 720) = x^3 + 720x^2 = 8^2 \cdot 8^2 \cdot 4^2 \cdot 81^2$. Let us now make $x = 8y$; this supposition gives $8^3y^3 + 720 \cdot 8^2y^2 = 8^2 \cdot 8^2 \cdot 4^2 \cdot 81^2$; and dividing by $8^3$, we have $y^3 + 90y^2 = 8 \cdot 4^2 \cdot 81^2$. Farther, let us suppose $y = 2z$, and we shall have $8z^3 + 4 \cdot 90z^2 = 8 \cdot 4^2 \cdot 81^2$; or, dividing by 8, $z^3 + 45z^2 = 4^2 \cdot 81^2$. Again, make $z = 9u$, in order to have, in this last equation, $9^3u^3 + 45 \cdot 9^2u^2 = 4^2 \cdot 9^4$, because dividing now by $9^3$, the equation becomes $u^3 + 5u^2 = 4^2 \cdot 9$, or $u^2(u + 5) = 16 \cdot 9 = 144$; where it is obvious, that $u=4$; for in this case $u^2 = 16$, and $u + 5 = 9$: since, therefore, $u = 4$, we have $z = 36$, $y = 72$, and $x = 576$, which is the less of the two numbers sought; so that the greater is $1296$, and the square root of this last, or $36$, multiplied by the other number $576$, gives $20736$. 730. *Remark*. This question admits of a simpler solution; for since the square root of the greater number, multiplied by the less, must give a product equal to a given number, the greater of the two numbers must be a square. If, therefore, from this consideration, we suppose it to be $x^2$, the other number will be $x^2 - 720$, which being multiplied by the square root of the greater, or by $x$, we have $x^3 - 720x = 20736 = 64 \cdot 27 \cdot 12$. If we make $x = 4y$, we shall have $64y^3 - 720 \cdot 4y = 64 \cdot 27 \cdot 12$, or $y^3 - 45y = 27 \cdot 12$. Supposing, farther, $y = 3z$, we find $27z^3 - 135z = 27 \cdot 12$; or, dividing by $27$, $z^3 - 5z = 12$, or $z^3 - 5z - 12 = 0$. The divisors of $12$ are $1, 2, 3, 4, 6, 12$: the first two are too small; but the supposition of $z = 3$ gives exactly $27 - 15 - 12 = 0$. Consequently, $z = 3$, $y = 9$, and $x = 36$; whence we conclude, that the greater of the two numbers sought, or $x^2$, $= 1296$, and that the less, or $x^2 - 720$, $= 576$, as before. 731. *Question 4*. There are two numbers, whose difference is $12$; and the product of this difference by the sum of their cubes is $102144$. What are the numbers? Calling the less of the two numbers $x$, the greater will be $x + 12$: also the cube of the first is $x^3$, and of the second S 2