341. If $x = \cos\alpha + i\sin\alpha, y = \cos\beta + i\sin\beta$, show that $\frac{(x + y)(xy - 1)}{(x - y)(xy + 1)} = \frac{\sin\alpha + \sin\beta}{\sin\alpha - \sin\beta}$.

Solution: L.H.S. $= \frac{xy(x + y) - (x + y)}{xy(x - y)+(x - y)}$. Dividing numerator and denominator by $xy$ $= \frac{x + y - \frac{1}{x} - \frac{1}{y}}{x - y + \frac{1}{y} - \frac{1}{x}}$

$= \frac{\cos\alpha + i \sin\alpha + \cos\beta + i \sin\beta - \cos\alpha + i\sin\alpha - \cos\beta + i \sin\beta}{\cos\alpha + i \sin\alpha - \cos\beta - i \sin\beta - \cos\alpha + i\sin\alpha + \cos\beta - i \sin\beta}$ $= \frac{\sin\alpha + \sin\beta}{\sin\alpha - \sin\beta} =$ R.H.S.

342. Show that ${}^nC_0 + {}^nC_3 + {}^nC_6 + \cdots = \frac{1}{3}\left[2^n + 2\cos\frac{n\pi}{3}\right]$.

Solution: $(1 + x)^n = {^nC_0} + {^nC_1}x + {^nC_3}x^2 + {^nC_3}x^3 + \cdots$

We know that $\omega, \omega^2 = \frac{-1\pm\sqrt{3}i}{2} = -cos\frac{\pi}{3} \pm \sin\frac{\pi}{3}$.

Putting $x = 1, \omega, \omega^2$ and adding we get

$2^n + 2\cos\frac{n\pi}{3} = 3[{}^nC_0 + {}^nC_3 + {}^nC_6 + \cdots] \Rightarrow {}^nC_0 + {}^nC_3 + {}^nC_6 + \cdots = \frac{1}{3}\left(2^n + 2\cos\frac{n\pi}{3}\right)$.

343. Show that ${}^nC_1 + {}^nC_4 + {}^nC_7 + \cdots = \frac{1}{3}\left[2^{n - 2} + 2\cos\frac{(n - 2)\pi}{3}\right]$.

Solution: Proceeding like previous question, $2^n = {}^nC_0 + {}^nC_1 + {}^nC_2 + {}^nC_3 + {}^nC_4 + {}^nC_5 + \cdots$

$(-\omega^2)^n = {}^nC_0 + {}^nC_1\omega + {}^nC_2\omega^2 + {}^nC_3\omega^3 + {}^nC_4\omega^4 + {}^nC_5\omega^5 + \cdots$

$\Rightarrow (-\omega^2)^{n}.\omega^2 = {}^nC_0\omega^2 + {}^nC_1\omega^3 + {}^nC_2\omega^4 + {}^nC_3\omega^5 + {}^nC_4\omega^6 + {}^nC_5\omega^7 + \cdots$

and $(-\omega)^n = {}^nC_0 + {}^nC_1\omega^2 + {}^nC_2\omega^4 + {}^nC_3\omega^6 + {}^nC_4\omega^8 + {}^nC_5\omega^{10} + \cdots$

$\Rightarrow (-\omega)^n.\omega = {}^nC_0\omega + {}^nC_1\omega^3 + {}^nC_2\omega^5 + {}^nC_3\omega^7 + {}^nC_4\omega^9 + {}^nC_5\omega^{11} + \cdots$

Adding $2^{n - 2} + 2\cos\frac{(n- 2)\pi}{3} = 3[{}^nC_1 + {}^nC_4 + {}^nC_7 + \cdots] \Rightarrow {}^nC_1 + {}^nC_4 + {}^nC_7 + \cdots = \frac{1}{3}\left[2^{n - 2} + 2\cos\frac{(n - 2)\pi}{3}\right]$

344. Show that ${}^nC_2 + {}^nC_5 + {}^nC_8 + \cdots = \frac{1}{3}\left[2^{n + 2} + 2\cos\frac{(n + 2)\pi}{3}\right]$.

Solution: This problem can be solved like previous problem. Put $x = 1, \omega, \omega^2$ and multiply with $1, \omega, \omega^2$ and then add to obtain the result.

345. If $C_r$ stands for ${}^{4n}C_r$, prove that $C_0 + C_4 + C_8 + \cdots = 2^{4n - 2} + (-1)^n2^{2n - 1}$.

Solution: $C_0 + C_1x + C_2x^2 + C_3x^3 + C_4x^4 + \cdots = (1 + x)^{4n}$. Putting $x = 1, -1, i, -i$ and adding

$4[C_0 + C_4 + C_8 + \cdots] = 2^{4n} + (1 + i)^{4n} + (1 - i)^{4n}$

$\Rightarrow C_0 + C_4 + C_8 + \cdots = 2^{4n - 2} + (-1)^n2^{2n - 1}$.

346. If $(1 - x + x^2)^{6n} = a_0 + a_1x + a_2x^2 + \cdots$, show that $a_0 + a_3 + a_6 + \ldots = \frac{1}{3}(2^{6n + 1} + 1)$.

Solution: Given $(1 - x + x^2)^{6n} = a_0 + a_1x + a_2x^2 + \cdots$. Putting $x = 1, \omega, \omega^2$

$1^{6n} = a_0 + a_1 + a_2 + a_3 + \cdots$

$(-2\omega)^{6n}= 2^{6n} = a_0 + a_1\omega + a_2\omega^2 + a_3\omega^3 + \cdots$

$(-2\omega^2)^{6n}= 2^{6n} = a_0 + a_1\omega^2 + _2\omega^4 + a_3\omega^6 + \cdots$

Adding $2^{6n + 1} + 1 = 3[a_0 + a_3 + a_6 + \cdots]$ $\Rightarrow a_0 + a_3 + a_6 + \cdots = \frac{1}{3}[2^{6n + 1} + 1]$.

347. If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \cdots$, show that $a_0 + a_3 + a_6 + \ldots = \frac{1}{3}\left(1 + (-1)^n2^{n + 1}\cos\frac{n\pi}{3}\right)$.

Solution: Proceeding like previous problem we obtain $3[a_0 + a_3 + a_6 + \cdots]$.

R.H.S. becomes $1^n + (-2\omega)^n + (-2\omega^2)^n$ but $-\omega = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$ and $-\omega^2 = \cos\frac{\pi}{3} - i\sin\frac{\pi}{3}$ and hence we have R.H.S.

348. Let $A = x + y + z, A' = x' + y' + z', AA' = x'' + y'' + z'', B = x + y\omega + z\omega^2, B' = x' + y'\omega + z'\omega^2, BB' = x'' + y''\omega + z''\omega^2, C = x + y\omega^2 + z\omega, C' = x' y'\omega^2 + z'\omega, CC' = x'' + y''\omega^2 + z''\omega,$ then find $x'', y''$ and $z''$ in terms of $x, y, z$ and $x', y' z'$.

Solution: Clearly, $x'' = \frac{AA' + BB' + CC'}{3}, y'' = \frac{AA' + BB'\omega^2 + CC'\omega}{3}$ and $z'' = \frac{AA' + BB'\omega + CC'\omega^2}{3}$,

and $AA' + BB' + CC' = (x + y + z)(x' + y' + z') + (x + y\omega + z\omega^2)(x' + y'\omega + z'\omega^2) + (x + y\omega^2 + z\omega)(x' + y'\omega^2 + z'\omega) = 3(xx' + zy' + yz')$. Analogously $y'' = yy' + zx' + xz', z'' = zz' + xy' + yz'$.

349. Prove the equaity $(ax - by - cz - dt)^2 + (bx + ay - dz + ct)^2 + (cx + dy + az - bt)^2 + (dx - cy + bz + at)^2 = (a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + t^2)$.

Solution: We have the identity $(\alpha\delta - \beta\gamma)(\alpha'\delta' - \beta'\gamma') = (\alpha\alpha' + \beta\gamma')(\gamma\beta' + \delta\delta') - (\alpha\beta' + \beta\delta')(\gamma\alpha' + \gamma\alpha' + \delta\gamma')$

Putting $\alpha = x + yi, \beta = z + ti, \gamma = -(z - ti), \delta = x - yi, \alpha' = a + bi, \beta' = c + di, \gamma' = -(c - di)$ and $\delta' = a - bi$ then

$\alpha\delta - \beta\gamma = x^2 + y^2 + z^2 + t^2$ and $\alpha'\delta' - \beta'\gamma' = a^2 + b^2 + c^2 + d^2$

$\Rightarrow \alpha\alpha' + \beta\gamma' = (ax - by - ca - dt) + i(bx + ay + dz - ct), \gamma\beta' + \delta\delta' = \overline{\beta\gamma'} + \overline{\alpha\alpha'} = \overline{\beta\gamma' + \alpha\alpha'}$

$\therefore \alpha\beta' + \beta\delta' = (cx - dy + az + bt) + i(dx + cy - bz + at), \gamma\alpha' + \delta\gamma' = -(cx - dy + az + bt) + i(dx + cy - bz + at)$

Thus, $-(\alpha\beta' + \beta\delta')(\gamma\alpha' + \delta\gamma') = (cx - dy + az + bt)^2 + (dx + cy - bz + at)^2$

Substituting obtained expression in the original idendity we have the required result.

350. Prove the equality: $\frac{\cos n\theta}{\cos^n\theta} = 1 - {}^nC_2\tan^2\theta + {}^nC_4\tan^4\theta - \ldots + A$, where $A = (-1)^{n/2}\tan^{n}\theta$ if $n$ is even, $A = (-1)^{(n - 1)/2}.{}^nC_{n - 1}\tan^n\theta$ if $n$ is odd.

Solution: $(\cos\theta + i\sin\theta)^n = \cos^n\theta + i.{}^nC_1\cos^{n - 1}\theta\sin\theta + i^2.{}^nC_2\cos^{(n - 2)}\theta\sin^2\theta + \cdots + i^r.{}^nC_r\cos{(n - r + 1)}\theta\sin^{r - 1}\theta + \cdots$

Separating real part, $\cos n\theta = \cos^n\theta - {}^nC_2\cos^{n - 2}\theta\sin^2\theta + \cdots$

Taking into account the parity of $n$ and dividing both members of these equalities by $\cos^n\theta$, we get the required formulas.