Q: How do you factor c cubed minus z cubed?

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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

Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265

z = 5

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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)

Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265

z = 5

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.

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

10/4 - z - 8/16 = 21/2 - z - 1/2 = 2 - z