Question

If x + y − z = 5, x 2 + y 2 + z 2 = 29; find the value of xy − yz − xz

Answer 1

Given

x+y+z=1,xy+yz+zx=−1 and xyz=−1

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

x3+y3+z3−3(−1)=1(x2+y2+z2)−(−1)

x3+y3+z3+3=x2+y2+z2+1

x3+y3+z3+2=x2+y2+z2

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

hope it helps you....

Answer 2

Step-by-step explanation:

✯Given:-

x + y - z = 5

x² + y² + z² = 29

✯To find:-

xy - yz - xz

✯Solution:-

We know that ,

(x+y-z)² = x²+y²+z²+ 2(xy-yz-xz)

Using this identity,

(5)² = 29 + 2(xy-yz-xz)

25 = 29 + 2(xy-yz-xz)

2(xy-yz-xz) = 25-29

2(xy-yz-xz)= -4

xy-yz-xz = -4 ÷ 2

xy-yz-xz=-2

xy-yz-xz= -2

Hope it helps

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More info:-

Some identities :-

• (a+b)² = a²+b²+2ab
• (a+b)²=(a+b)(a+b)
• (a-b)²=(a-b)(a-b)
• a²+b² = (a+b)² - 2ab
• a²+b² = (a-b)² + 2ab
• a²-b² = (a-b)(a+b)
• (a+b+c)² = a²+b²+c²+2(ab+bc+ca)
• (a+b)³ = a³+b³+3ab(a+b)
• a³+b³ = (a+b)(a²+b²-ab)
• (a-b)³ = a³-b³-3ab(a-b)
• a³-b³ = (a-b)(a²+b²+ab)