Question

if a3+b3=35 and a+b=5 then find the value of 1/a+1/b

Answer 1

✍✍HERE IS YOUR ANSWER..⬇⬇
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$\underline{SOLUTION}$ ➡

$\underline{Given \: : }\: \: a {}^{3} + b {}^{3} = 35 \: - - - - > (1)\\ \\ a + b = 5 \: - - - - > (2)\\ \\ \\ \underline{To \: \: find \: : }\: \: \frac{1}{a} + \frac{1}{b} =?$

$We \: \: know \: \: that , \\ \\ a {}^{3} + b {}^{3} = ( a+b )(a {}^{2} - ab+ b {}^{2} ) \\ \\ = > a {}^{3} + b {}^{3} = ( a + b)[(a + b) {}^{2} - 3 ab] \\ \\ = > 35 = 5[(5) {}^{2} - 3ab]\: \: \: \: [from \: \: (1) \: \: and \: \: (2)]\\ \\ = > 35 = 5(25 - 3ab) \\ \\ = > 35 = 125 - 15ab \\ \\ = > 125 - 15ab = 35 \\ \\ = > - 15ab = 35 - 125 \\ \\ = > - 15ab = - 90 \\ \\ = > ab = \frac{ - 90}{ - 15} \\ \\ = > ab = 6 \: - - - - > (3)$

$Now, \\ \\ \frac{1}{a} + \frac{1}{b} \\ \\ = \frac{b + a}{ab} \\ \\ = \frac{a + b}{ab} \\ \\ = \frac{5}{6} \: \: \: \: [from \: (1) \: \: and \: (3)]$

$\underline{ANSWER}$ ➡ $\boxed{\bold{\frac{1}{a} + \frac{1}{b} = \frac{5}{6}}}$

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✝✝…HOPE…IT…HELPS…YOU…✝✝

Answer 2

Here is your answer buddy,

Given ,

$A^{3} + B^{3}$ = 35 ----------(1)

A+B = 5 ----------(2)

Now,

We know that,

$A^{3} + B^{3}$ =$(A+B)^{3}$ - 3AB(A+B)

=> 35 = $5^{3}$ - 3AB(5)

=> 35 = 125 - 15AB (since, Cube of 5 = 125 )

=> 35 - 125 = -15AB

=> - 90 = -15AB

=> AB = 6

Now, as per question, we have to find the value of $\frac{1}{A} + \frac{1}{B}$

So,

$\frac{1}{A} + \frac{1}{B}$

=> $\frac{A+B}{AB}$

=> $\frac{5}{6}$

Hope this helps you.
Be Brainly.