Question

If 2x-3y-4z =0 ,then find 8x^3-27y^3-64z^3​

Answer 1

Answer:

The correct answer will be = (-72xyz)

Step-by-step explanation:

As we know the formula a^3+b^3+c^3-3abc=(a+b+c)*(a^2+b^2+c^2-ab-bc-ca).

So rewrite the question to form like given the above formula i_e.

[(2x)^3+(-3y)^3+(-4y)^3]=[(2x)+(-3y)+(-4z)]*{[(2x)^2+(-3y)^2+(-4z)^2]-(2x)(-3y)-(-3y)(-4z)-(-4z)(2x)}-3(2x)(-3y)(-4z)

as 2x-3y-4z=0, putting in the above equation so we get the following result.

(2x-3y-4z)*(4x^2+9y^2+16z^2+6xy-12yz+8zx)-72xyz.

(0)*(4x^2+9y^2+16z^2+6xy-12yz+8zx) - 72xyz = -72xyz.

Hope you find it helpful :)

Answer 2

Answer:

$8 {x}^{3} - 27 {y}^{3} - 64 {z}^{3} = 72xyz$

Step-by-step explanation:

as we know that

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

but this formula has a special case ,

$if \: x + y + z = 0 \\ \\ then \\ \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

Now ATQ,

$2x - 3y - 4z = 0$

let us try to convert the given expression according to the formula

$8 {x}^{3} - 27 {y}^{3} - 64 {z}^{3} \\ \\ {(2x)}^{3 } + {( - 3y)}^{3} + {( - 4z)}^{3}$

Given that

$2x - 3y - 4z = 0 \\ \\ so \\ \: 8 {x}^{3} - 27 {y}^{3} - 64 {z}^{3} = 3(2x)( - 3y)( - 4z) \\ \\ 8 {x}^{3} - 27 {y}^{3} - 64 {z}^{3} = 72xyz$

Hope it helps you.