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I’m done with it all but imma keep my promise and give 100 to the person who can solve this one for my challenge question on here

Im done with it all but imma keep my promise and give 100 to the person who can solve this one for my challenge question on here class=

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Ankit

Answer:

[tex]\boxed{{a}^{4} + {b}^{4} + {c}^{4} =50}[/tex]

Step-by-step explanation:

[tex]a+b+c = 4[/tex]

Squaring both the side,

[tex]{(\underline{a +b} + c)}^{2} = {4}^{2} [/tex]

[tex]{(a + b)}^{2} + {c}^{2} + 2(a + b)c \: = 16[/tex]

[tex] {a}^{2} + 2ab+ {b}^{2} + {c}^{2} + 2ac + 2bc = 16[/tex]

[tex]{a}^{2} + {b}^{2} + {c}^{2} + 2ab+ 2ac + 2bc = 16[/tex]

[tex]10 + 2ab+ 2ac + 2bc = 16[/tex]

[tex] 2ab+ 2ac + 2bc = 16 - 10 = 6[/tex]

[tex]2(ab + bc + ac) = 6[/tex]

[tex] \fbox{ab + bc + ac = 3}[/tex]

Now we know that,

[tex] {a}^{3} + {b}^{3} + {c}^{3} -3abc = ( a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)[/tex]

[tex]22 -3abc = ( 4)( 10 - (ab + bc + ca))[/tex]

[tex]22 -3abc = ( 4)( 10 - 3)[/tex]

[tex]3abc = 22 - 40 + 12[/tex]

[tex]abc = \frac{ - 6}{3} = - 2[/tex]

Now, we know that

[tex]ab + bc + ac = 3[/tex]

Squaring both the side,

[tex] {(ab + bc + ac)}^{2} = {3}^{2} [/tex]

[tex] {(ab)}^{2} + {(ac)}^{2} + {(bc)}^{2} + 2 (abbc + abac + bcac) = 9[/tex]

[tex]{(ab)}^{2} + {(ac)}^{2} + {(bc)}^{2} + 2 (a + b + c)(abc) = 9 [/tex]

[tex]{(ab)}^{2} + {(ac)}^{2} + {(bc)}^{2} + 2 \times 4 \times ( - 2) = 9[/tex]

[tex]{(ab)}^{2} + {(ac)}^{2} + {(bc)}^{2} - 16= 9[/tex]

[tex]{(ab)}^{2} + {(ac)}^{2} + {(bc)}^{2} = 25[/tex]

Now we are given that,

[tex] {a}^{2} + {b}^{2} + {c}^{2} = 10[/tex]

Squaring both the side,

[tex]{({a}^{2} + {b}^{2} + {c}^{2})}^{2} = {10}^{2} [/tex]

[tex] { {a}^{2} }^{2} + { {b}^{2} }^{2} + { {c}^{2} }^{2} + 2( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {a}^{2} {c}^{2} ) = 100[/tex]

[tex] {a}^{4} + {b}^{4} + {c}^{4} + 2( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {a}^{2} {c}^{2} ) = 100[/tex]

[tex]{a}^{4} + {b}^{4} + {c}^{4} + 2( 25 ) = 100[/tex]

[tex]{a}^{4} + {b}^{4} + {c}^{4} + 50= 100 [/tex]

[tex] \boxed{{a}^{4} + {b}^{4} + {c}^{4} =50}[/tex]

[tex] \small \sf Thanks \: for \: joining \: brainly \: community![/tex]

MisterBrainly

There is a secret rule hiding inbetween these equations.If we can spot out the rule we won't have to do lengthy calculations .we can do it in one line even then .

First note that degree of equation is taken as d for all the context below provided

When

d=1

  • The expression yields 4

d=2

  • expression yeilds=10

d=3

  • expression yields=22

d=4 we need what the expression yields

See if we observe

For d=1

  • The expression equals =4

Next for d=2 and d=3

The respective yields are 10 and 22

Spot the secret

  • 4²-6=10
  • 4²+6=22

70% secret exposed,rest we do

Find the difference

  • 22-10=12

Then

  • 6 is 4×1+2

Next one must be 4×12+2 (As it's difference before)

It's exactly

  • 48+2=50

So the next chain for d=4 and d=5 are

  • (d=2)²±50
  • 10²±50
  • 50 and 150

So the respective values are

  • a⁴+b⁴+c⁴=50

  • a⁵+b⁵+c⁵=150

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