Question

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

Answer 1

Answer: 72xyz

Step-by-step explanation:

Use the formula

If $a + b + c = 0$ then $a^{3} + b^{3} + c^{3} = 3abc$

Proof

a+b+c = 0

as we know,

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² -ab-bc-ac)

a³ + b³ + c³ -3abc = (0)(a²+b²+c²-ab-bc-ac)

a³+b³+c³ -3abc = 0

a³+b³+c³ = 3abc

Over here, 2x + (-3y) + (-4z) = 0

Then $(2x)^{3} + (-3y)^{3} + (-4z)^{3} = 3*(2x)*(-3y)*(-4z)$

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

Please brainlist my answer, if helpful!

Answer 2

Answer:

Step-by-step explanation:

Use the formula

If a + b + c = 0a+b+c=0 then a^{3} + b^{3} + c^{3} = 3abca

3

+b

3

+c

3

=3abc

Proof

a+b+c = 0

as we know,

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² -ab-bc-ac)

a³ + b³ + c³ -3abc = (0)(a²+b²+c²-ab-bc-ac)

a³+b³+c³ -3abc = 0

a³+b³+c³ = 3abc

Over here, 2x + (-3y) + (-4z) = 0

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

3

+(−3y)

3

+(−4z)

3

=3∗(2x)∗(−3y)∗(−4z)

=> 8x^{3} - 27y^{3} - 64z^{3} = 72xyz8x

3

−27y

3

−64z

3

=72xyz

Please brainlist my answer, if helpful!