It can't be factored because its discriminant is less than zero.
4x cubed y cubed z divided by x negative squared y negative 1 z sqaured = 4
(z-9)(z+3
z = 5
Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265
This is a difference of two sqaures. 49y2 - z2 = (7y - z)(7y + z)
It can't be factored because its discriminant is less than zero.
4x cubed y cubed z divided by x negative squared y negative 1 z sqaured = 4
The only common factor to all terms is yz. → xy³z² + y²z + xyz = yz(xy²z + y + x)
(z-9)(z+3
ax + ay - AZ = a(x + y - z)
z = 5
Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265
In Roman numerals L and C are 50 and 100 respectively. So substitute 50 and 100 into the equation and let the unknown be called z. 50-z = 100 -z = 100-50 -z = 50 --z = -50 Solution: z = -50 Substituting -50 into the equation: 50--50 = 100 or L--L = C. Remember that in maths a double minus sign before a number changes that number into a plus number. Check it out on your calculator. David Gambell, Merseyside, England.
The equation x^3 + y^3 = z^3 is known as Fermat's Last Theorem, which states that there are no integer solutions for x, y, and z when the exponent is greater than 2. This theorem was famously proven by mathematician Andrew Wiles in 1994 after centuries of attempts. Therefore, there are no whole number solutions to the equation x^3 + y^3 = z^3.
10/4 - z - 8/16 = 21/2 - z - 1/2 = 2 - z
Im not sure if you can type the numbers smaller so just know that all the numbers in this answer should be powers. x4+y2-z3