If x^{1/3} + y^{1/3} = z^{1/3}, then (x + y - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $x^{1/3} + y^{1/3} = z^{1/3}$ ......$(1)$Cube both sides of the equation:$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$Using the identity $(a + b)^3

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given: $x^{1/3} + y^{1/3} = z^{1/3}$ ......$(1)$Cube both sides of the equation:$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:$x + y + 3x^{1/3}y^{1/3}(x^{1/3} + y^{1/3}) = z$Substitute $x^{1/3} + y^{1/3} = z^{1/3}$ from equation $(1)$:$x + y + 3x^{1/3}y^{1/3}z^{1/3} = z$Rearrange the terms:$x + y - z = -3x^{1/3}y^{1/3}z^{1/3}$Cube both sides again:$(x + y - z)^3 = (-3x^{1/3}y^{1/3}z^{1/3})^3$$(x + y - z)^3 = -27xyz$Therefore:$(x + y - z)^3 + 27xyz = 0$

If $x^{1/3} + y^{1/3} = z^{1/3},$ then $(x + y - z)^3 + 27xyz$ is equal to

Similar Questions

$7^{5} \div 7^{3} \div 7^{2} \times 7^{4} \times \frac{1}{7^{3}} = 7^{?}$

The value of $3+\frac{3}{3+\frac{1}{3+\frac{1}{3}}}$ is:

If $1$ is added to the denominator of a fraction,the fraction becomes $\frac{1}{2}$. If $1$ is added to the numerator of the fraction,the fraction becomes $1$. The fraction is

$\sqrt{5^{2} \times 14 - 6 \times 7 + (4)^{?}} = 18$

$34.95 + 240.016 + 23.98 = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module