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.
There are an infinite number of real solutions. For example, x = 1, y = 1 and z = cuberoot(2), or x = 1, y = 2 and z = cuberoot(9). There are no integer solutions, as proven by Fermat's Last Theorem.
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