Question

if x+y+z=5 and xy+yz+xz=10, then find the value of x^3+y^3+z^3-3xyz.​

Answer 1

Answer:

Solution(By Examveda Team)

Givenx+y+z=6xy+yz+zx=10To find x3+y3+z3−3xyz = ?⇒ Using formula,

⇒(x+y+z)2=

x2+

y2+

z2+

2(xy+yz+zx)

⇒62=x2+y2+z2+2×10⇒36=x2+y2+z2+20⇒x2+y2+z2=16

⇒x2+y2+z2−3xyz=

(x+y+z)

(x2+y2+z2−xy−yz−zx)

⇒x2+y2+z2−3xyz=6[16−10]⇒x2+y2+z2−3xyz=6×6⇒x2+y2+z2−3xyz=36

Answer 2

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find the value of x³ + y³ + z³ - 3xyz

$\huge\bf\underline \purple{ \mathcal{S}olution:-}$

$\bf\underline{\red{Given:-}}$

$\bullet \rm \: x + y + z = 5$

$\bullet \rm \: xy + yz + xz = 10$

• We know that,

$\rm \: x {}^{3} + {y}^{3} + {z}^{3} - 3xyz : -$

$\red{ \rm = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) }$

= 5[x² + y² + z² -(10)] ....1)

• Finding value of x² + y² + z²

we know that,

✫ (x + y + z)² = x² + y² + z² + 2xyb + 2yz + 2xz

↛ x² + y² + z² = (x + y + z)² -( 2xyb + 2yz + 2xz)

↛x² + y² + z² = 5² -2(10)

↛x² + y² + z² = 25 - 20

↛ x² + y² + z² = 5

• putting value of x² + y² + z² in 1)

Then,

= 5[x² + y² + z² -(10)]

= 5[5 -10]

= 5( -5)

= -25

Hence,

• Value of x³ + y³ + z³ - 3xyz is -25

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