x to the power 13 * y to the power 14 *(xy)to the power 5 *(xyz)to the power 12 / (xyz)to the power 4*(x)to the power10
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Step-by-step explanation:
a²+b⁴ = 47
Explanation:
Given: For a, b are real numbers such that a+b=3 and a^2+b^2=7a
2
+b
2
=7 .
Using Identities: a^4+b^4=(a^2+b^2)^2-2a^2\cdot b^2a
4
+b
4
=(a
2
+b
2
)
2
−2a
2
⋅b
2
[1]
First find the value of a^2b^2a
2
b
2
.
(a+b)^2=a^2+b^2+2a\cdot b(a+b)
2
=a
2
+b
2
+2a⋅b [2]
From the given condition, we have a+b=3 and a^2+b^2=7a
2
+b
2
=7 .
Substitute these values in equation [2], we have
(3)^2=7+2ab(3)
2
=7+2ab or
9=7+2ab
Simplify:
ab=1 or
a^2 \cdot b^2=1a
2
⋅b
2
=1
Now, to substitute the value of a^2 \cdot b^2=1a
2
⋅b
2
=1 and a^2+b^2=7a
2
+b
2
=7 in equation [1];
a^4+b^4=(7)^2-2\cdot 1=49-2=47a
4
+b
4
=(7)
2
−2⋅1=49−2=47
therefore, the value of a^4+b^4= 47a
4
+b
4
=47
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