Question

a + b + c = 4\
a² + b² + c² = 10\
a³ + b³ + c³ = 22\
a⁴ + b⁴ + c⁴ = ?\
hardest maths questions - find\
a^4 + b^4 + c^4\
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a^x + b^y = c^z \
a + 2^b + 2^c = \
= ? b = ?\
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ow to solve th\
+ b + c = a\
b + b + b =\
b × b × b =\
2b⁵ × 4b² =

Explanation:

Step1: Find \( ab + bc + ca \)

We know that \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \). Substituting the given values \( a + b + c = 4 \) and \( a^2 + b^2 + c^2 = 10 \), we get:
\[
4^2 = 10 + 2(ab + bc + ca)
\]
\[
16 = 10 + 2(ab + bc + ca)
\]
Subtract 10 from both sides:
\[
6 = 2(ab + bc + ca)
\]
Divide by 2:
\[
ab + bc + ca = 3
\]

Step2: Find \( abc \)

We use the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \). Substituting \( a + b + c = 4 \), \( a^2 + b^2 + c^2 = 10 \), \( ab + bc + ca = 3 \), and \( a^3 + b^3 + c^3 = 22 \):
\[
22 - 3abc = 4(10 - 3)
\]
\[
22 - 3abc = 4 \times 7
\]
\[
22 - 3abc = 28
\]
Subtract 22 from both sides:
\[
-3abc = 6
\]
Divide by -3:
\[
abc = -2
\]

Step3: Find \( a^4 + b^4 + c^4 \)

We use the identity \( (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) \). First, find \( a^2b^2 + b^2c^2 + c^2a^2 \). We know that \( (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) \). Substituting \( ab + bc + ca = 3 \), \( abc = -2 \), and \( a + b + c = 4 \):
\[
3^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2(-2)(4)
\]
\[
9 = a^2b^2 + b^2c^2 + c^2a^2 - 16
\]
Add 16 to both sides:
\[
a^2b^2 + b^2c^2 + c^2a^2 = 25
\]

Now, substitute into the identity for \( (a^2 + b^2 + c^2)^2 \):
\[
10^2 = a^4 + b^4 + c^4 + 2 \times 25
\]
\[
100 = a^4 + b^4 + c^4 + 50
\]
Subtract 50 from both sides:
\[
a^4 + b^4 + c^4 = 50
\]

Answer:

\( 50 \)