Sum of Fifth Powers

Expanding $(a+b)^3$ we have $(a+b)^3=a^3+b^3+3ab(a+b) => ab=-1$

Expanding $(a+b)^5$ we have $(a+b)^5=a^5+b^5+5ab(a^3+b^3)+10(ab)^2(a+b) => a^5+b^5=82$

$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$

$\implies 14 = 2^3 - 3ab \times 2$ [putting the value of $(a+b)$ and $(a^3 + b^3)$ ]

$\implies 14 - 8 = - 6ab$

$\implies 6 = -6ab$

$\implies ab = -1$

$(a + b)^5 \implies a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + a^5$

So, $a^5 + b^5 \implies (a+b)^5 - (5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4)$

$\implies (a+b)^5 - [( 5a^4b + 5ab^4 ) + ( 10a^3b^2 + 10a^2b^3)]$

$\implies 2^5 - [{5ab(a^3 + b^3)} + {10a^2b^2(a+b)}]$

$\implies 32 - [(5 \times -1 (a^3 + b^3)) + (10 \times (-1)^2 (a+b))]$

$\implies 32 - (( -5 \times 14) + (10 \times 2))$

$\implies 32 - (-70 + 20)$

$\implies 32 - (-50) \implies 32 + 50 \implies \boxed{82}$

Given $a+b=2 , a^{3}+b^{3}=14$ , and applying , * $a^{3}+b^{3}=(a+b)^{3} - 3ab(a+b)$ , we get $ab= -1$ *

We know that $a^{2}+b^{2} = (a+b)^{2} - 2ab = 6$ (In our sum) .

Now * $(a^{2}+b^{2})(a^{3}+b^{3})=a^{5}+b^{5}+(ab)^{2}(a+b)$ , giving $a^{5}+b^{5}= 82...$ *

We observe, $(a^{3}+b^{3})(a^{2}+b^{2})=a^{5}+b^{5}+a^{3}b^{2}+a^{2}b^{3}$ $=a^{5}+b^{5}+a^{2}b^{2}(a+b)$

Also, $(a+b)^{2}=a^{2}+b^{2}+2ab$ $\implies$ $4-2ab=a^{2}+b^{2}$ $-----(\$)$

Now, $14=a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)=2(4-3ab)$ [From $(\$)$ ]

$\implies$ $ab=-1.$

Therefore, $a^{2}+b^{2}=4-2ab=4-2(-1)=6.$

Hence, $(a^{3}+b^{3})(a^{2}+b^{2})=14 \times 6=84.$

Therefore, $a^{5}+b^{5}=(a^{3}+b^{3})(a^{2}+b^{2})-a^{2}b^{2}(a+b)=84-(-1)^{2} \times 2=\boxed{82}$ .

Factorizing the expression we want to evaluate and using $a+b=2$ :

$a^5 + b^5 = (a+b)(a^4 - a^3b +a^2b^2 - ab^3 + a^4) = 2(a^4 - a^3b +a^2b^2 - ab^3 + a^4) \hspace{2 mm} (I)$

Now we need to find the value of the other factor. Factorizing $a^3+b^3$ :

$a^3+b^3 = (a+b)(a^2-ab+b^2) = 2(a^2-ab+b^2) = 14$

$a^2-ab+b^2 = 7 \hspace{2 mm} (II)$

Squaring both sides of $(II)$ and manipulating:

$a^4 + (ab)^2 + b^4 - 2a^3b + 2a^2b^2-2ab^3 = 49$

$a^4 - a^3b + a^2b^2 - ab^3 + a^4 = 49 - 2a^2b^2 + a^3b + ab^3$ $= 49 - 2(ab)^2 + ab(a^2 + b^2) \hspace{2 mm} (III)$

What is $ab$ and $a^2+b^2$ ? Squaring both sides of $a+b = 2$ , we get that $a^2 + 2ab+b^2 = 4$ .

From this and $(II)$ we obtain $ab = -1$ and $a^2+b^2 = 6$ . Plugging that into $(III)$ :

$a^4 - a^3b + a^2b^2 - ab^3 + a^4 =49 - 2(-1)^2 + (-1)(6) = 41$

Substituting back into $(I)$ :

$a^5 + b^5 = 2 \cdot 41 = \boxed{82}$

First of all, let's square the both sides of the equation $a+b=2$ , and we will use it later on.

$4 = (a+b)^{2}$

$4 = a^{2} + 2ab +b^{2}-(1)$

Now, let's factorize $a^{3} + b^{3} = 14$ .

$14 = (a+b)(a^{2} - ab + b^{2}$

Substitute the equation $a+b = 2$ into the equation above, we get

$14 = 2(a^{2} - ab + b^{2}$

Divide $2$ on both sides,

$7 = a^{2} - ab + b^{2} - (2)$

$(2) - (1)$ ,

$3 = -3ab$

$ab = -1$

Substitute $ab = -1$ into $(2)$ ,

$7 = a^{2} + 1 b^{2}$

$6 = a^{2} + b^{2}$

Then,

$(a+b)^{4} = 16$

$a^{4} + b^{4} + 4ab(b^{3} + a^{3}) + 6(ab)^{2} = 16$

Substitute the equation $a^{3} + b^{3} = 14$ and $ab = -1$ into the equation above,

a^{4} + b^{4} -4(14} + 6 = 16

$a^{4} + b^{4} = 66$

Then,

$(a+b)^{5} = a^{5} +b^{5} +5ab(a^{3} + b^{3}) + 10(ab)^{2}(a+b)$

Substitute the equation $a^{3} + b^{3} = 14$ and $ab = -1$ into the equation above,

$32 = a^{5} +b^{5} - 5(14) +10(2)$

So, we get $a^{5} + b^{5} = \boxed {82}$ in the end.

a+b = 2 a^3 + b^3 = 14

Now, a^3+b^3 = (a + b) * (a^2 + b^2 - ab) Therefore, a^2 + b^2 - ab = 7 ...................................(1)

Also, (a + b)^2 = a^2 + b^2 + 2ab Hence, 4 = a^2 + b^2 + 2ab.......................................(2)

From (1) and (2) , ab = -1..........................................(3)

Using (3) and a+b=2 , we solve for a and b using the quadratic formula. We find, a = 1 + root(2) or 1 - root(2) b = 1 - root(2) or 1 + root(2)

Substituting these values in the required expression, we find, a^5 + b^5 = 82

a+b=2

a^3+b^3=14= (a+b)(a^2-ab+b^2)

a^2-ab+b^2=7

a^2+2ab+b^2=4

then we substitute the first equation to the second equation

ab=-1

a-1/a-2=0 b-1/b-2=0

then we substitute tha answer into a^5+b^5 henced we got82

$a^3+b^3=14$ , we can factorize this to give $(a+b)(a^2-ab+b^2)=14$ , but $a+b=2$ , so $a^2-ab+b^2=7$ .

$a+b=2$ , so $a=2-b$ , we can substitute this into $a^2-ab+b^2=7$ to give $(2-b)^2-b(2-b)+b^2=7$ . Solving this quadratic tells us that $b=1\pm \sqrt{2}$ and $a=1\mp \sqrt{2}$ , we can then just substitute these values into $a^5+b^5$ to get our answer of $82$ .

(a+b)^3= a^3+b^3+ 3ab(a+b) putting a+b = 2 and a^3+b^3=14 we get, ab=-1 and a(2-a)= -1 gives quadratic equation a^2-2a-1=0, splving this we get a = 1+ \sqrt{2} and b= 1-\sqrt{2} . Putting these values of a and b in a^5+b^5, we get the answer 82

Observe that $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$ , so let's obtain first $ab$ with the known values:

$14=(2)^{3}-3ab(2) \leftrightarrow ab=-1$

Now, observe that $a^{5}+b^{5}=(a+b)^{5}-5ab(a^{3}+b^{3}) - 10(ab)^{2}(a+b)$ , so replace with the known values again:

$a^{5}+b^{5}=(2)^{5}-5(-1)(14)-10(-1)^{2}(2)$

$a^{5}+b^{5}=\boxed{82}$

a+b=2

a^3+b^3=(a+b)(a^2-xy+b^2)

from the equation we get (a^2-xy+b^2)=7

a^2+2ab+b^2=4 (from the 1st line) then solve with (a^2-xy+b^2)=7 you'll get ab=-1

solve ab=-1 with a+b=2 you'll get if

a=1-sqrt2 b=1+sqrt2,

if a=1+sqrt2 b=1-sqrt2

just insert it to f(a,b)=a^5+b^5 youll get 82

Observe $(a+b)^3=a^3+b^3+3ab(a+b)=8\implies ab=-1$ by plugging in what we know. Now $(a+b)^2=a^2+b^2+2ab=a^2+b^2-2=4$ so $a^2+b^2=6$ . Finally $84=6*14=(a^2+b^2)(a^3+b^3)=a^5+b^5+a^2b^2(a+b)=a^5+b^5+2$ hence $a^5+b^5=84-2=\boxed{82}$ .

Let $T_n=a^n+b^n$ Then: $T_{n+1}=a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1})=(a+b)T_n-abT_{n-1}\\ \boxed{T_{n+1}=2T_n-abT_{n-1}}$ Hence we have found a recurrence relation!!!.Now all we need is the value of $ab$ .This can easily be found as follows: $a^3+b^3=(a+b)(a^2-ab+b^2)\\=(a+b)((a+b)^2-3ab)=14\\2(4-3ab)=14\\ \rightarrow -3ab=7-4\rightarrow ab=-1$ Hence: $T_{n+1}=2T_n+T_{n+1}$ The rest is simple computation: $T_3=2T_2+T_1\rightarrow \boxed{T_2=6}\\ T_4=2T_3+T_2\rightarrow \boxed{T_4=34}\\ T_5=2T_4+T_3\rightarrow \boxed{T_5=82}$

Using binomial theorem,

$(a+b)^{5} = C{5\choose 0} a^5 b^0 + C{5\choose 1} a^4 b^1 + C{5\choose 2} a^3 b^2 + C{5\choose 3} a^2 b^3 + C{5\choose 4} a^1 b^4 + C{5\choose 5} a^0 b^5$

which is equal to: $a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a^{1} b^{4} + b^{5}$ Leave it here only.............

Now get to the problem,

We know that: $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$

putting values we get: $8 = 14 + 6ab$

On solving we get $\boxed{ab = -1}$

Now get back to where we started.....

We had: $(a+b)^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a^{1} b^{4} + b^{5}$

But this can be written as: $(a+b)^{5} = a^{5} + (5)(ab) (a)^{3} + 10a (ab)^{2} + 10b (ab)^{2} + 5(ab) b^{3} + b^{5}$

Taking commons : $(a+b)^{5} = a^{5} + (5)(ab) ( a^{3} + b^{3} ) + 10(a+b) (ab)^{2} + b^{5}$

Putting values of $(a+b) , ( a^{3} + b^{3} ), ab$ ,we get :

$(2)^{5} = a^{5} - (5)*(14)+ (10)(2) + b^{5}$

which is equal to: $32 = a^{5} - 70 + 20 + b^{5}$

Thus $a^{5} + b^{5}$ is equal : $\boxed{82}$

(a+b)^5 = a^5 + 5 a^ 4b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5
2^5 = a^5 + b^5 + 5ab (a^3 + b^3) + 10(ab)^2[a+b]
32 = a^5 +b^5 +70ab + 20(ab)^2
How to find the value of ab?
Using (a+b)^3 = a^3 + 3ab(a+b) + b^3 , you can easly find that ab= -1
So: 32 = a^5 + b^5 + 20 - 70
a^5 + b^5 = 82

a + b = 2 a^{3} + b^{3} = 14

(a+b)(a^{2} + b^{2} + 2ab ) = 2^{3} ( a^{3} + b^{3} )+ ab^{2} + 2a^{2} b + ba^{2} +2ab^{2} = 8 14 + 3 ( ab^{2} + ba^{2} ) = 8 ab^{2} + ba^{2} = -2 ab( a + b ) = -2 ab = -1 a = -1 / b

a + b = 2 -1/b + b = 2 -1 + b^(2} -2b = 0 b = 1±\sqrt{2} a = -1 / 1±\sqrt{2}

Substitute a = -1 / 1±\sqrt{2} and b = 1±\sqrt{2} into a^{5} + b^{5} , Hence, answer = 82