Answer:
[tex]\Huge\boxed{ \bf \dfrac{ {b}^{2} + {a}^{2} }{ {a}^{3}b }} [/tex]
Step-by-step explanation:
Given expression:
[tex] \sf \longmapsto \dfrac{ {a}^{3} {b}^{5} }{b {}^{4} {a}^{6} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
Solution:
[tex] \sf \longmapsto \dfrac{ {a}^{3} {b}^{5} }{b {}^{4} {a}^{6} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
Cancel a^3 and a^6 on the left hand side of plus sign, which results to a^3.
[tex] \sf \longmapsto\dfrac{ \cancel{{a}^{3}} {b}^{5} }{b {}^{4} \cancel{{a}^{6}} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
That is,
[tex] \sf \longmapsto \dfrac{b {}^{5} }{b {}^{4} a {}^{3} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
Cancel b^5 and 5^4 on the LHS of plus sign, which results to b.
[tex]\sf \longmapsto \dfrac {\cancel{b {}^{5} }}{ \cancel{b {}^{4}} a {}^{3} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
That is,
[tex]\sf \longmapsto \dfrac{b }{ a {}^{3} } + \dfrac{ {a}^{8} {b}^{6} }{b {}^{7} a {}^{9} } [/tex]
Now cancel a^8 and a^9 on the RHS of Plus sign, which results to a.
[tex] \sf \longmapsto\dfrac{b }{ a {}^{3} } + \dfrac{ \cancel{{a}^{8}} {b}^{6} }{b {}^{7} \cancel{a {}^{9}} }[/tex]
That is,
[tex] \sf \longmapsto \: \dfrac{b }{ a {}^{3} } + \dfrac{ {b}^{6} }{b {}^{7} a {}^{} }[/tex]
Cancel b^6 and b^7 on the RHS of the Plus sign, which results to b.
[tex] \sf \longmapsto\dfrac{b }{ a {}^{3} } + \dfrac{ \cancel{ {b}^{6}} }{ \cancel{b {}^{7}} a {}^{} }[/tex]
That is,
[tex]\sf \longmapsto\dfrac{b }{ a {}^{3} } + \cfrac {1}{ {b {}^{1}} a {}^{} }[/tex]
Simply add:-
Rewrite into:
[tex] \sf \longmapsto \: \dfrac{ b {}^{} }{a {}^{3} } + \cfrac{1}{a \times b} [/tex]
Combine the numerators over LCD(a^3)
[tex] \sf \longmapsto \: \dfrac{ {b}^{2} + {a}^{2} }{ {a}^{3}b } [/tex]
Or it can also be rewritten as,
[tex] \sf \longmapsto \: \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{3}b } [/tex]
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I hope this helps!
Please let me know if you have any questions
