Question

a=√6+√5,p+q=6,pq=3 then find the
value of p5+q5=4806​

Answer 1

Answer:

I hope you can take help from this process

Step-by-step explanation:

Using the identity is (a+b)3=a3+b3+3ab(a+b)

Consider,

(p+q)3=p3+q3+3pq(p+q)

 

It is given that p+q=5 and pq=6, therefore,

(p+q)3=p3+q3+3pq(p+q)⇒(5)3=p3+q3+(3×6)(5)⇒125=p3+q3+(18×5)⇒125=p3+q3+90⇒p3+q3=125−90⇒p3+q3=35

Hence, p3+q3=35.

Answer 2

Answer:

4086

Step-by-step explanation:

Using the given data and binomial theorem, we get,

$a = \sqrt{6} + \sqrt{5} \\ p + q = 6 \: \: \: pq = 3$

${(p + q)}^{5} = \binom{5}{0} \times {p}^{5} + \binom{5}{1} \times {p}^{4} \times q + \binom{5}{2} \times {p}^{3} \times {q}^{2} + \binom{5}{3} \times {p}^{2} \times {q}^{3} + \binom{5}{4} \times p \times {q}^{4} + \binom{5}{5} \times {q}^{5} \\ = {p}^{5} + 5 {p}^{4}q + 10 {p}^{3} {q}^{2} + 10 {p}^{2} {q}^{3} + 5p {q}^{4} + {q}^{5} = ( {p}^{5} + {q}^{5} ) + 5pq( {p}^{3} + 2 {p}^{2}q + 2p {q}^{2} + {q}^{3} ) \\ = > {6}^{5} = ( {p}^{5} + {q}^{5} ) + 5 \times3 \times (( {p}^{3} + {q}^{3} ) + 2pq(p + q)) \\ = > {p}^{5} + {q}^{5} = {6}^{5} - 15 (( {p}^{3} + {q}^{3} ) + 2 \times 3 \times 6)$

Now, we have to find p^3+q^3. To do this, we apply binomial theorem with index 3 or the commnoly know as (a+b)^3 formula.

${(p + q)}^{3} = {p}^{3} + 3 {p}^{2} q + 3p {q}^{2} + {q}^{3} \\ = ( {p}^{3} + {q}^{3} ) + 3pq(p + q) \\ = > {6}^{3} = {p}^{3} + {q}^{3} + 3 \times 3 \times 6 \\ = > {p}^{3} + {q}^{3} = {6}^{3} - 6 \times 9 = 216 - 54 = 162$

Inserting this value in the initial equation we get,

${p}^{5} + {q}^{5} = {6}^{5} - 15 \times (162 + 36) \\ = {6}^{5} - 15 \times 198 = 7776 - 2970 \\ = > {p}^{5} + {q}^{5} = 4806$

Hope this helps you. PLEASE MARK ME AS THE BRAINLIEST.