Question

If x+y+z=1,xyz=-1and xy+yz+xz=-1 then find the the value of x³+y³+z³.

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\fbox{Formula\;to\;be\;used}$

x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² × yz × zx × xy)

${\boxed{\sf\:{We\;can\;also\;write\;it}}}$

x³ + y³ + z³ = (x + y + z)(x² + y² + z² × yz × zx × xy) + 3xyz

$\fbox{Put\;the\;values}$

x³ + y³ + z³= (1) {(x²+y²+ z²- (-1)} + 3(-1)

x³ + y³ + z³= (1)(x² + y² + z² + 1) - 3 ....... (1)

$\fbox{Therefore}$

x² + y² + z²

(x + y + z)² = x² + y² + z² + 2 (xy + yz + zx)

(1)² = x² + y² + z² + 2(-1)

1 = x² + y² + z² - 2

1 + 2 = x² + y² + z²

3 = x² + y² + z² ........ (2)

${\boxed{\sf\:{Substitute\;the\;value\;of\;(2)\;in\;(1)}}}$

x³ + y³ + z³= (1)(3 + 1) - 3

x³ + y³ + z³= (1)(4) - 3

x³ + y³ + z³ = 1

$\boxed{\begin{minipage}{9 cm} Additional Information \\ \\ \ 1)(x+y)^2=x^2+y^2+2xy \\ \\ 2)(x-y)^2=x^2+y^2-2xy \\ \\ 3)(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx \\ \\ 4)(x+y)^3=x^3+y^3+3xy(x+y) \\ \\ 5)(x-y)^3=x^3-y^3-3xy(x-y) \end{minipage}}$

Answer 2

hope it helps mark as brainliest please follow me