History of Hindu Mathematics: A Source Book v2.c17.s24.21

v2.c17.s24.21

“The sum of their squares = 2y² - 2yu² + u⁴. This is a cube. Making it equal to u⁶ and transposing we get u⁶ - u⁴ = 2y² - 2yu². Multiplying both sides by 2 and superadding u⁴, we get the square-root of the second side = 2y - u², and the first side = 2u⁶ - u⁴. Dividing out by u⁴ (and putting w for 2y/u² - 1), we get 2u² - 1 = w². By the method of the Square-nature the roots of this equation are u &= 5, 29, &\dots / w &= 7, 41, &\dots "Then by the rule, 'Or, if a biquadratic factor has been removed, the greater root should be multiplied by the square of the lesser root,'¹ we get