Question

If x+y+z=3,x^2+y^2+z^2=5,x^3+y^3+z^3=7,find the value of x^4+y^4+z^4

Answer 1

Given :-

• x + y + z = 3
• x^2+y^2+z^2=5
• x^3+y^3+z^3=7.

To Find :-

• x^4+y^4+z^4 = ?

Solution :-

Let ,

→ x + y + z = 3 ---------- Equation (1)

→ x^2+y^2+z^2=5 --------- Equation (2)

→ x^3+y^3+z^3=7 ----------- Equation (3) .

Squaring Both Sides of Equation (1) Now, we get,

→ (x + y + z)² = 3²

→ x² + y² + z² + 2(xy + yz + zx) = 9

Putting value of Equation (2) now, in LHS,

→ 5 + 2(xy + yz + zx) = 9

→ 2(xy + yz + zx) = 9 - 5

→ (xy + yz + zx) = 2 . -------------- Equation (4).

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Now, using x ³ + y ³ + z ³ - 3xyz = (x + y + z) (x ² + y ² + z²-xy -xz- yz)

→x ³ + y ³ + z ³ - 3xyz = (x + y + z) (x ² + y ² + z²-xy -xz- yz)

→ x ³ + y ³ + z ³ - 3xyz = (x + y + z)[x ² + y ² + z²- (xy+xz+yz)]

Putting values of Equation (1) , (2), (3) & Equation (4) in This formula we get,

→ 7 - 3xyz = 3 [ 5 - 2 ]

→ 7 - 3xyz = 3 * 3

→ 3xyz = 7 - 9

→ 3xyz = (-2)

→ xyz = (-2/3). ----------------------- Equation (5).

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Now,

☛ (x⁴ + y⁴ + z⁴) = (x² + y² + z²)² - 2(x².y²+y².z²+x².z²)

Or,

☛ (x⁴ + y⁴ + z⁴) = (x² + y² + z²)² - 2[(xy+yz+xz)² - 2xyz(x+y+z)]

Putting Our values of Equation (2) , Equation (4), Equation (5) & Equation (1) in RHS Respectively, we get,

☞ (x⁴ + y⁴ + z⁴) = (5)² - 2[ 2² - 2 * (-2/3) * 3 ]

☞ (x⁴ + y⁴ + z⁴) = 25 - 2[ 4 + 4 ]

☞ (x⁴ + y⁴ + z⁴) = 25 - 16

☞ (x⁴ + y⁴ + z⁴) = 9 (Ans).

(Nice Question.)

Answer 2

Given,

• $\sf{x+y+z=3\quad\quad\dots(1)}$

• $\sf{x^2+y^2+z^2=5\quad\quad\dots(2)}$

• $\sf{x^3+y^3+z^3=7\quad\quad\dots(3)}$

We know that,

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

Thus we obtain,

$\longrightarrow\sf{(x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)\quad\quad\dots(4)}$

And

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

$\longrightarrow\sf{(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)\quad\quad\dots(5)}$

From (4), we have,

$\longrightarrow\sf{x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)\quad\quad\dots(6)}$

And from (5), we have,

$\longrightarrow\sf{x^2y^2+y^2z^2+z^2x^2=(xy+yz+zx)^2-2xyz(x+y+z)\quad\quad\dots(7)}$

Substituting (7) in (6),

$\longrightarrow\sf{x^4+y^4+z^4=(x^2+y^2+z^2)^2-2((xy+yz+zx)^2-2xyz(x+y+z))\ \dots(8)}$

We have to find the values of $\sf{xy+yz+zx}$ and $\sf{xyz.}$

Squaring (1),

$\longrightarrow\sf{(x+y+z)^2=3^2}$

$\longrightarrow\sf{x^2+y^2+z^2+2(xy+yz+zx)=9}$

Substituting (2),

$\longrightarrow\sf{5+2(xy+yz+zx)=9}$

$\longrightarrow\sf{xy+yz+zx=2}$

We know that,

$\longrightarrow\sf{(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))=x^3+y^3+z^3-3xyz}$

$\longrightarrow\sf{3(5-2)=7-3xyz}$

$\longrightarrow\sf{xyz=-\dfrac{2}{3}}$

Then (8) becomes,

$\longrightarrow\sf{x^4+y^4+z^4=5^2-2(2^2+2\times\dfrac{2}{3}\times3)}$

$\longrightarrow\sf{\underline{\underline{x^4+y^4+z^4=9}}}$

Hence 9 is the answer.