without using the tables, find the value of (2+√3)^6 +(2-√3)^6
I see the sum of cubes
(2+√3)^6 +(2-√3)^6
= [(2+√3)^2]^3 + [(2-√3)^2]^3
= (4 + 4√3 + 3)^3 + (4 - 4√3) + 3)^3
= (7 + 4√3)^3 + (7 - 4√3)^3 , now from a^3 + b^3= (a+b)(a^2 - ab + b^2)
= (7+4√3 + 7-4√3)( (2+√3)^4 - (2+√3)^2 (2-√3)^2 + (2-√3)^4 )
look at the middle term, (2+√3)^2 (2-√3)^2
[(2+√3)(2-√3)]^2 = [4 - 3]^2 = 1 , nice!
for the first and last terms, expand it using the binomial expansion, things will cancel
= 14(.....) , believe it or not, that last part comes out to a whole number, I will let you work it out
let me know what your final answer is, I have written mine down
Well, solving that equation without using tables can be quite a challenge. It's like trying to juggle flaming bowling pins blindfolded while riding a unicycle. But fear not, my friend! I am here to offer you a little mathematical clown magic.
Now, let's dive right into it. We can use a little trick called the binomial theorem to simplify this expression. According to this theorem, the sixth power of a binomial can be expanded as follows:
(n + m)^6 = C(6, 0)n^6m^0 + C(6, 1)n^5m^1 + C(6, 2)n^4m^2 + C(6, 3)n^3m^3 + C(6, 4)n^2m^4 + C(6, 5)n^1m^5 + C(6, 6)n^0m^6
Where C(n, k) represents the binomial coefficient "n choose k".
Now, let's substitute n = 2 and m = √3 into this equation:
(2 + √3)^6 = C(6, 0)(2)^6(√3)^0 + C(6, 1)(2)^5(√3)^1 + 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)(2)^0(√3)^6
Now, since we are aiming for value rather than a literal representation, bear with me while I simplify these terms.
C(6, 0)(2)^6(√3)^0 = 1(2)^6(1) = 64
C(6, 1)(2)^5(√3)^1 = 6(2)^5(√3) = 192√3
C(6, 2)(2)^4(√3)^2 = 15(2)^4(√3)^2 = 720
C(6, 3)(2)^3(√3)^3 = 20(2)^3(√3)^3 = 960√3
C(6, 4)(2)^2(√3)^4 = 15(2)^2(√3)^4 = 720
C(6, 5)(2)^1(√3)^5 = 6(2)^1(√3)^5 = 192√3
C(6, 6)(2)^0(√3)^6 = 1(√3)^6 = 729
Now, let's combine all the simplified terms:
(2 + √3)^6 = 64 + 192√3 + 720 + 960√3 + 720 + 192√3 + 729
Now, we can group the numeric constants and the square root terms:
(2 + √3)^6 = (64 + 720 + 720 + 729) + (192√3 + 960√3 + 192√3)
Adding the numeric constants separately:
(64 + 720 + 720 + 729) = 2173
Adding the square root terms separately:
(192√3 + 960√3 + 192√3) = 1344√3
Now adding both results:
(2 + √3)^6 = 2173 + 1344√3
Voila! We've got our answer without using tables. However, if you have a table on hand, I must admit it would have made the calculation much easier. But hey, where's the fun in that?
To find the value of (2+√3)^6 + (2-√3)^6 without using tables, we can use the binomial theorem. The binomial theorem states that the expansion of (a + b)^n can be found by summing the terms of the form C(n, k) * a^k * b^(n-k), where C(n, k) represents the binomial coefficient and is equal to n! / (k! * (n-k)!), n is the exponent, k is the index of the term, and a and b are the numbers being raised to the power.
In this case, we have (2+√3)^6 + (2-√3)^6, so we can use the binomial theorem to expand each term separately. Let's start with the term (2+√3)^6:
(2+√3)^6 = C(6, 0) * (2^6) * (√3^0) + C(6, 1) * (2^5) * (√3^1) + 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) * (2^0) * (√3^6)
Now let's simplify each term:
C(6, 0) * (2^6) * (√3^0) = 1 * 2^6 * 1 = 64
C(6, 1) * (2^5) * (√3^1) = 6 * 2^5 * √3 = 192√3
C(6, 2) * (2^4) * (√3^2) = 15 * 2^4 * 3 = 720
C(6, 3) * (2^3) * (√3^3) = 20 * 2^3 * 3√3 = 240√3
C(6, 4) * (2^2) * (√3^4) = 15 * 2^2 * 9 = 540
C(6, 5) * (2^1) * (√3^5) = 6 * 2 * 27√3 = 324√3
C(6, 6) * (2^0) * (√3^6) = 1 * 1 * 81 = 81
Now let's add up all the terms: 64 + 192√3 + 720 + 240√3 + 540 + 324√3 + 81.
Simplifying the terms with √3, we get: 64 + 720 + 540 + 81 + (192 + 240 + 324)√3.
Combining like terms, we have: 1405 + 756√3.
Now let's move on to the second term, (2-√3)^6:
Using the same steps, we can expand (2-√3)^6 and simplify each term:
C(6, 0) * (2^6) * (√3^0) = 64
C(6, 1) * (2^5) * (√3^1) = -192√3
C(6, 2) * (2^4) * (√3^2) = 720
C(6, 3) * (2^3) * (√3^3) = -240√3
C(6, 4) * (2^2) * (√3^4) = 540
C(6, 5) * (2^1) * (√3^5) = -324√3
C(6, 6) * (2^0) * (√3^6) = 81
Adding up all the terms: 64 - 192√3 + 720 - 240√3 + 540 - 324√3 + 81.
Simplifying the terms with √3, we get: 64 + 720 + 540 + 81 + (-192 - 240 - 324)√3.
Combining like terms, we have: 1405 - 756√3.
Finally, we can find the sum of (2+√3)^6 + (2-√3)^6:
(2+√3)^6 + (2-√3)^6 = (1405 + 756√3) + (1405 - 756√3) = 1405 + 1405 = 2810.
Therefore, the value of (2+√3)^6 + (2-√3)^6 is 2810.
To find the value of (2+√3)^6 +(2-√3)^6 without using tables, we can use the binomial expansion formula. The formula states that for any binomial expression (a + b)^n, the expansion can be determined by summing the terms of the form (n choose k) * a^(n-k) * b^k, where (n choose k) represents the binomial coefficient and is equal to n! / (k!(n-k)!).
In this case, we want to find the value of (2+√3)^6 +(2-√3)^6. We can treat (2+√3) and (2-√3) as the "a" and "b" terms, respectively.
Let's expand (2+√3)^6:
(2+√3)^6 = (6 choose 0) * 2^6 * (√3)^0 + (6 choose 1) * 2^5 * (√3)^1 + (6 choose 2) * 2^4 * (√3)^2 + (6 choose 3) * 2^3 * (√3)^3 + (6 choose 4) * 2^2 * (√3)^4 + (6 choose 5) * 2^1 * (√3)^5 + (6 choose 6) * 2^0 * (√3)^6
We can simplify this expression by evaluating the binomial coefficients and exponents:
= 1 * 2^6 * 1 + 6 * 2^5 * √3 + 15 * 2^4 * (3) + 20 * 2^3 * (3√3) + 15 * 2^2 * (9) + 6 * 2^1 * (9√3) + 1 * 2^0 * (27)
= 64 + 192√3 + 240 + 160√3 + 60 + 12√3 + 27
= 334 + 364√3
Now let's expand (2-√3)^6 using the same process:
(2-√3)^6 = (6 choose 0) * 2^6 * (√3)^0 + (6 choose 1) * 2^5 * (√3)^1 + (6 choose 2) * 2^4 * (√3)^2 + (6 choose 3) * 2^3 * (√3)^3 + (6 choose 4) * 2^2 * (√3)^4 + (6 choose 5) * 2^1 * (√3)^5 + (6 choose 6) * 2^0 * (√3)^6
= 1 * 2^6 * 1 + 6 * 2^5 * -√3 + 15 * 2^4 * (3) + 20 * 2^3 * (-3√3) + 15 * 2^2 * (9) + 6 * 2^1 * (-9√3) + 1 * 2^0 * (27)
= 64 - 192√3 + 240 - 160√3 + 60 - 12√3 + 27
= 334 - 364√3
Finally, we can find the value of (2+√3)^6 +(2-√3)^6:
(2+√3)^6 +(2-√3)^6 = (334 + 364√3) + (334 - 364√3)
The terms with √3 will cancel each other out, resulting in:
= 334 + 334
= 668
Therefore, (2+√3)^6 +(2-√3)^6 = 668.