Question

Without using table find the value of
(2+root3)^6+(2-root3)^6

Answer 1

To find the value of \( (2+\sqrt{3})^6 + (2-\sqrt{3})^6 \), we can use the binomial theorem and properties of symmetric polynomials.

Let us denote:
\[
x = 2 + \sqrt{3}
\]
\[
y = 2 - \sqrt{3}
\]
We want to calculate \( x^6 + y^6 \).

Notice that:
\[
x + y = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4
\]
\[
xy = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1
\]

Now, we can use a recursive relationship derived from the properties of symmetric sums. The sequence defined by \( a_n = x^n + y^n \) satisfies the recurrence relation:
\[
a_n = (x+y)a_{n-1} - xy a_{n-2}
\]

For our specific case, we can initialize:
\[
a_0 = x^0 + y^0 = 2
\]
\[
a_1 = x + y = 4
\]

Now applying the recurrence relation:
\[
a_2 = (x+y)a_1 - xy a_0 = 4 \cdot 4 - 1 \cdot 2 = 16 - 2 = 14
\]
\[
a_3 = (x+y)a_2 - xy a_1 = 4 \cdot 14 - 1 \cdot 4 = 56 - 4 = 52
\]
\[
a_4 = (x+y)a_3 - xy a_2 = 4 \cdot 52 - 1 \cdot 14 = 208 - 14 = 194
\]
\[
a_5 = (x+y)a_4 - xy a_3 = 4 \cdot 194 - 1 \cdot 52 = 776 - 52 = 724
\]
\[
a_6 = (x+y)a_5 - xy a_4 = 4 \cdot 724 - 1 \cdot 194 = 2896 - 194 = 2702
\]

Thus, the value of \( (2+\sqrt{3})^6 + (2-\sqrt{3})^6 \) is:
\[
\boxed{2702}
\]