Question

If x+y+z=12 and x^2+y^2+z^2=70,then find the value of x^3+y^3+z^3-3xyz

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

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

x + y + z = 12

x² + y² + z² = 70

★We know identity :-

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

★Substitute the values get :-

(12)² = 70 + 2(xy + yz + zx)

144 = 70 + 2(xy + yz + zx)

144 - 70 = 2(xy + yz + zx)

74 = 2(xy + yz + zx)

$\tt{\rightarrow\dfrac{74}{2}=xy+yz+zx}$

xy + yz + zx = 37

★Here we have,

x³ + y³ + z³ - 3xyz

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

= (x + y + z)[(x² + y² + z² - (xy + yz + zx)]

★We get the values above :-

= (12)(70 - 37)

= 12 × 33

= 396

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