Question

____________________________​

Answer 1

Answer:

Given

• a + b + c = 3 ___________ ( 1 )
• a² + b² + c² = 9 _________ ( 2 )
• a³ + b³ + c³ = 12 ________ ( 3 )

To find:

• a⁴ + b⁴ + c⁴ = 0

Solution:

Taking equation 1

a + b + c = 3

Squaring on both sides

(a + b + c)² = 3²

(a + b + c)² = 9 ___________ ( 4 )

We know that

2(ab + bc + ac) = (a + b + c)² - (a² + b² + c²)

2(ab + bc + ac) = 3² - 9

(°.° Equation 4 & 2)

2(ab + bc + ac) = 0

ab + bc + ac = 0__________ ( 5 )

Taking equation 3

a³ + b³ + c³ = 12

Adding + 3abc - 3abc on LHS

a³ + b³ + c³ + 3abc - 3abc = 12

(a + b + c)(a² + b² + c² - ab - bc - ac) + 3abc = 12

(a + b + c)[a² + b² + c² - (ab + bc + ac)] + 3abc = 12

Substituting the values of equation 1 , 2 & 5

(3)(9 - 0) + 3abc = 12

12 - 3(9) = 3abc

12 - 27 = 3abc

- 15 = 3abc

- 15/3 = 6abc

abc = - 5

Now,

a²b² + b²c² + a²c² = (ab + bc + ac)² - 2abc(a + b + c)

★ 0 - 2(-5)(3)

★ 30 ________ ( 6 )

Finding : a⁴ + b⁴ + c⁴

Using

x⁴ + y⁴ + z⁴ = (x + y + z)⁴ - 2(x²y² + y²z² + x²z²)

→ a⁴ + b⁴ + c⁴ = [(a + b + c)²]² - 2{[(ab)² + (bc)² + (ac)² + 2(ab + bc + ac)]}

→ a⁴ + b⁴ + c⁴ = 9² - 2[a²b² + b²c² + a²c² + 2(0)]

→ a⁴ + b⁴ + c⁴ = 81 - 2(a²b² +b²c² + a²c² + 0)

→ a⁴ + b⁴ + c⁴ = 81 - 2(30)

(°.° Equation 6)

→ a⁴ + b⁴ + c⁴ = 81 - 60

→ a⁴ + b⁴ + c⁴ = 21

Hence, a⁴ + b⁴ + c⁴ = 21

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{a + b + c = 3} \\ &\sf{ {a}^{2} + {b}^{2} + {c}^{2} = 9}\\ &\sf{ {a}^{3} + {b}^{3} + { c}^{3} = 12 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{ {a}^{4} + {b}^{4} + {c}^{4} = ? }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\large \underline{\tt \: \red{ According \: to \: statement }}$

$\longmapsto\tt \: a + b + c = 3$

On squaring both sides, we get

$\longmapsto\tt \: {(a + b + c)}^{2} = {3}^{2}$

$\longmapsto\tt \: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 9$

$\longmapsto\tt \: 9 + 2(ab + bc + ca) = 9$

$\longmapsto\tt \: 2(ab + bc + ca) =0$

$\rm :\implies\: \pink{ab + bc + ca=0}$

Now,

• We know that

$\longmapsto\tt \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc \: =$

$\tt \: (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca))$

On substituting the values, we get

$\rm :\implies\:12 - 3abc = 3(9 - 0)$

$\rm :\implies\:3abc = 12 - 27$

$\rm :\implies\:3abc = - 15$

$\rm :\implies\: \pink{abc = - 5}$

Now,

Consider,

$\tt \: (ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)$

On substituting the above evaluated values and given values, we get

$\rm :\implies\: {0}^{2} = {a}^{2} {b}^{2} + {b}^{2}{c}^{2} +{c}^{2}{a}^{2} + 2( - 5)(3)$

$\rm :\implies\:0 = {a}^{2} {b}^{2} + {b}^{2}{c}^{2} +{c}^{2}{a}^{2} - 30$

$\bf\implies \: \red{{a}^{2} {b}^{2} + {b}^{2}{c}^{2} +{c}^{2}{a}^{2} = 30}$

Now,

We know that,

$\longmapsto\tt \: \tt \: {( {a}^{2}+ {b}^{2}+ {c}^{2} )}^{2}$

$\tt \: = {a}^{4}+{b}^{4} +{c}^{4} + 2({a}^{2} {b}^{2} + {b}^{2}{c}^{2} +{c}^{2}{a}^{2})$

On Substituting the value from above evaluated and given values, we get

$\rm :\implies\: {(9)}^{2} = {a}^{4}+{b}^{4} +{c}^{4} + 2(30)$

$\rm :\implies\:81 = {a}^{4}+{b}^{4} +{c}^{4} + 60$

$\bf\implies \: \blue{{a}^{4}+{b}^{4} +{c}^{4} = 21}$