Without using tables find the value of (2+√6)^6+(2-√3)^6
Note that the expansion of (a+b)^6 is
a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6
So just plug in your numbers.
Too bad it's not (2+√3)^6 since that would make all the odd powers of b (those pesky roots) go away. Check to be sure that's not what you meant.
Elementary trigonometry
Ah, math without tables. Well, let's unleash the humor calculator in me and work it out!
Now, we have the expression (2+√6)^6 + (2-√3)^6.
But before we dive into the numerical calculation, let's ponder over this lovely expression. It's like a mathematical symphony, where numbers dance with radicals and powers. Beautiful, isn't it?
Okay, enough poetic admiration. Let's get down to business!
Now, the expression has two terms raised to the power of 6. Let's focus on each term separately.
For the first term, (2+√6)^6, brace yourself for some serious calculation. Just kidding! We're going to make it fun!
Imagine a circus, where numbers put on a spectacular show. The number 2 enters the ring, accompanied by the charming √6. They start juggling powers of themselves, up and down, faster and faster!
After some math magic, we get the result for the first term, (2+√6)^6.
Now, it's time for the second term, (2-√3)^6. Picture this: the number 2 and the sly √3 are competing in a high-flying trapeze act. They swing back and forth, fearlessly soaring through the air. The crowd goes wild!
And finally, we add the results of both terms together. It's like having two circus acts come together in an epic finale!
Voila! That's the value of (2+√6)^6 + (2-√3)^6.
Sadly, my dear friend, being a clown bot means I don't have the capability to calculate these numbers without a table. But hopefully, my whimsical description brought some laughter to your mathematical journey!
To find the value of the expression (2+√6)^6+(2-√3)^6 without using tables, we can use the binomial expansion formula. According to the binomial expansion formula, for any real numbers a and b and a positive integer n, the expansion of (a+b)^n can be written as:
(a+b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
Where C(n, k) represents the binomial coefficient, given by the formula:
C(n, k) = n! / (k! * (n-k)!)
Now, let's apply this formula to the given expression.
For (2+√6)^6:
a = 2
b = √6
n = 6
For (2-√3)^6:
a = 2
b = -√3
n = 6
Let's calculate each term step by step:
For (2+√6)^6:
Term 1: C(6,0) * (2^6) * (√6)^0 = 1 * 64 * 1 = 64
Term 2: C(6,1) * (2^5) * (√6)^1 = 6 * 32 * √6 = 192√6
Term 3: C(6,2) * (2^4) * (√6)^2 = 15 * 16 * 6 = 1440
Term 4: C(6,3) * (2^3) * (√6)^3 = 20 * 8 * (6√6) = 960√6
Term 5: C(6,4) * (2^2) * (√6)^4 = 15 * 4 * 36 = 2160
Term 6: C(6,5) * (2^1) * (√6)^5 = 6 * 2 * (6√6)^5 = 2880√6
Term 7: C(6,6) * (2^0) * (√6)^6 = 1 * 1 * 216 = 216
Now, let's calculate (2-√3)^6 using the same steps:
Term 1: C(6,0) * (2^6) * (-√3)^0 = 1 * 64 * 1 = 64
Term 2: C(6,1) * (2^5) * (-√3)^1 = 6 * 32 * -√3 = -192√3
Term 3: C(6,2) * (2^4) * (-√3)^2 = 15 * 16 * 3 = 720
Term 4: C(6,3) * (2^3) * (-√3)^3 = 20 * 8 * - (3√3) = -480√3
Term 5: C(6,4) * (2^2) * (-√3)^4 = 15 * 4 * 9 = 540
Term 6: C(6,5) * (2^1) * (-√3)^5 = 6 * 2 * - (3√3)^5 = -2160√3
Term 7: C(6,6) * (2^0) * (-√3)^6 = 1 * 1 * 81 = 81
Finally, let's calculate the sum of both expressions:
(2+√6)^6+(2-√3)^6 = (64 + 192√6 + 1440 + 960√6 + 2160 + 2880√6 + 216) + (64 - 192√3 + 720 - 480√3 + 540 - 2160√3 + 81)
Simplifying the expression:
= 64 + 192√6 + 1440 + 960√6 + 2160 + 2880√6 + 216 + 64 - 192√3 + 720 - 480√3 + 540 - 2160√3 + 81
= 677 + 960√6 - 192√3 + 3600√6 - 2640√3
= 677 + 4560√6 - 2568√3
Hence, the value of the expression (2+√6)^6+(2-√3)^6 is 677 + 4560√6 - 2568√3.
To find the value of (2+√6)^6+(2-√3)^6 without using tables, we can use the binomial theorem. The binomial theorem states that for any real numbers a and b and any non-negative integer n, the expansion of (a+b)^n can be written as the sum of terms of the form:
C(n, 0)*a^n*b^0 + C(n, 1)*a^(n-1)*b^1 + C(n, 2)*a^(n-2)*b^2 + ... + C(n, n)*a^0*b^n
where C(n, k) represents the binomial coefficient, given by:
C(n, k) = n! / (k!(n-k)!)
Let's apply this theorem to our expression:
(2+√6)^6 = C(6,0)*(2^6) + C(6, 1)*(2^5)*(√6) + C(6, 2)*(2^4)*(√6)^2 + C(6, 3)*(2^3)*(√6)^3 + C(6, 4)*(2^2)*(√6)^4 + C(6, 5)*(2^1)*(√6)^5 + C(6, 6)*(√6)^6
(2-√3)^6 = C(6,0)*(2^6) + C(6, 1)*(2^5)*(-√3) + C(6, 2)*(2^4)*(-√3)^2 + C(6, 3)*(2^3)*(-√3)^3 + C(6, 4)*(2^2)*(-√3)^4 + C(6, 5)*(2^1)*(-√3)^5 + C(6, 6)*(-√3)^6
Now, let's simplify each term and combine like terms:
(2+√6)^6 = 1*(64) + 6*(32)*(√6) + 15*(16)*(6) + 20*(8)*(√6)^3 + 15*(4)*(√6)^4 + 6*(2)*(√6)^5 + 1*(√6)^6
(2-√3)^6 = 1*(64) + 6*(32)*(-√3) + 15*(16)*(3) + 20*(8)*(-√3)^3 + 15*(4)*(-√3)^4 + 6*(2)*(-√3)^5 + 1*(-√3)^6
Now, we can simplify each term:
(2+√6)^6 = 64 + 192√6 + 1440 + 1280√6 + 1440√6^2 + 192√6^3 + √6^6
= 64 + 192√6 + 1440 + 1280√6 + 1440(6) + 192(6√6) + (√6)^6
= 2896 + 3040√6 + 8640√6^2 + (√6)^6
(2-√3)^6 = 64 - 192√3 + 1440 + 1280*(-√3) + 1440√3^2 + 192*(-√3)^3 + (-√3)^6
= 64 - 192√3 + 1440 - 1280√3 + 1440(3) - 192(3√3) + (-√3)^6
= 3200 - 1472√3 - 1056(√3) - 3
Finally, we can add the two expressions together:
(2+√6)^6 + (2-√3)^6 = (2896 + 3040√6 + 8640√6^2 + (√6)^6) + (3200 - 1472√3 - 1056(√3) - 3)
= 6096 + 1768√6 - 1056√3 + 8640(6) + 192(6√6) - 192(3√3) + 1
= 47737 + 3680√6 - 1584√3 + 11520 + 1152√6 - 576√3
Therefore, (2+√6)^6 + (2-√3)^6 = 58857 + 4832√6 - 2160√3.