Question

If x+y+z=16, and xy+yz+zx =11 find x2+y2+z2

Answer 1

Answer:

Step-by-step explanation:

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

 

⇒(x+y+z)2= ⇒ 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

Answer :

$\large{\star\:\:\boxed{\bf{x^2\:+\:y^2\:+\:z^2\:=\:234}}\:\:\star}$

Explanation :

Given :–

• x + y + z = 16
• xy + yz + zx = 11

To Find :–

• Value of x² + y² + z² = ?

Formula Used :–

• $\boxed{\bf{\star\:\:(a\:+\:b\:+\:c)^2\:=\:(a^2\:+\:b^2\:+\:c^2)\:+\:2(ab\:+\:bc\:+\:ca)\:\:\star}}$

Solution :–

We have x + y + z = 16 and xy + yz + zx = 11 .

Putting these values in the Formula :

$\rightarrow\sf{(a\:+\:b\:+\:c)^2\:=\:(a^2\:+\:b^2\:+\:c^2)\:+\:2(ab\:+\:bc\:+\:ca)}$

$\rightarrow\sf{(x\:+\:y\:+\:z)^2\:=\:(x^2\:+\:y^2\:+\:z^2)\:+\:2(xy\:+\:yz\:+\:zx)}$

$\rightarrow\sf{(16)^2\:=\:(x^2\:+\:y^2\:+\:z^2)\:+\:2(11)}$

$\rightarrow\sf{256\:=\:(x^2\:+\:y^2\:+\:z^2)\:+\:22}$

$\rightarrow\sf{x^2\:+\:y^2\:+\:z^2\:=\:256\:-\:22}$

$\rightarrow\boxed{\bf{x^2\:+\:y^2\:+\:z^2\:=\:234}}$

∴ The value of x² + y² + z² is 234 .

$\dag$ Some More Useful Formulae :-

→ (x + y)³ = x³ + y³ + 3xy(x + y)

→ (x - y)³ = x³ - y³ - 3xy(x - y)

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

  [Note : If x + y+ z = 0 then x³ + y³ + z³ = 3xyz .]