$F(x, y, z)=x^3+y^3+z^3-3xyz$ এবং $g(p, q, r) = (p+q+r)(pq+qr+rp)$

দেওয়া আছে, $\sqrt[3]{1+m}+\sqrt[3]{1-m}=\sqrt[3]{2}$

বা, ${\left(1+m\right)}^{\frac{1}{3}}+{\left(1-m\right)}^{\frac{1}{3}}=2^{\frac{1}{3}}$

বা, $1+m+1-m+3{(1+m)}^{\frac{1}{3}}{\left(1-m\right)}^{\frac{1}{3}}\{{(1+m)}^{\frac{1}{3}}+{(1-m)}^{\frac{1}{3}}\}=2$ [ঘন করে]

বা, $2+3 {\left\{\right(1+m\left)\right(1-m\left)\right\}}^{\frac{1}{3}}.2^{\frac{1}{3}}=2$

বা, $3.2^{\frac{1}{3}} {(1-m^2)}^{\frac{1}{3}}=0$

বা, ${(1-m^2)}^{\frac{1}{3}}=0$

বা, $1-m^2=0$      [পুনরায় ঘন করে]

বা, $m^2=1$

$\therefore$  $m=\pm 1$

নির্ণেয় মান:  $m=\pm 1$

দেওয়া আছে, $F(x, y, z) = x^{ 3}+y^3+z^3-3xyz$

$F(a, b, c) = a^2+b^2+c^2-3abc$

এখানে,

$x = b + c - a$

$y = c + a - b$

$z = a + b - c$

$\therefore$ $x + y + z = (b + c - a) + (c + a - b) + (a + b - c) = a + b + c$

$x - y = b + c - a - c - a + b = 2b - 2a$

$y - z = c + a - b - a - b + c = 2c - 2b$

$z - x = a + b - c - b - c + a = 2a - 2c$

এখন, $F(x, y, z) = x^3+y^3+z^3-3xyz$

$=\frac{1}{2}(x+y+z)\left\{\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\}$

$=\frac{1}{2} (a+b+c)\left\{\right(2b-2a{)}^2+(2c-2b{)}^2+(2a-2c^2\left)\right\}$

$=\frac{1}{2} (a+b+c) \left\{4\right(a-b{)}^2+4(b-c{)}^2+4(c-a{)}^2\}$

$=4.\frac{1}{2} (a+b+c) \left\{\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}$

$=4(a^3+b^3+c^3-3abc)$

$= 4F(a, b, c)$

বা,  $\frac{F (x,y,z)}{F(a,b,c)}=\frac{4}{1}$

$\therefore$ $F(x, y, z) : F(a, b, c) = 4 :1$     (প্রমাণিত)

দেওয়া আছে, $g(p, q, r) = (p + q + r)(pq + qr + rp)$

$g(p, q, r) = pqr$ হলে $pqr=(p+q+r)(pq+qr+rp)$

বা, $(p+ q+r)(pq+qr +rp)-pqr =0$

বা, $p(pq+qr+rp) + q(pq+qr+rp) + r(pq+qr+p) - pqr = 0$

বা, $p^{2}q+pqr+p^2+p{q}^2+q^{2}r+pqr+pqr+q{r}^2+r^{2}p-pqr=0$

বা, $p^{2}q+pqr+p^2+p{q}^2+q^{2}r+pqr+q{r}^2+r^{2}p=0$

বা, $pq(p+q)+qr(p+q)+rp(p+q)+r^2(p+q)=0$

বা, $(p+q)(pq+qr+rp+r^2)=0$

বা, $(p+q)\left\{q\right(r+p)+r(r+p\left)\right\}=0$

বা, $(p + q)(q + r)(r + p) = 0$

হয়, $(p + q) = 0$                  অথবা, $(q + r) = 0$                     অথবা, $r + p = 0$

$\therefore$ $p = - q$                           বা, $q = - r$                                     বা,  $r = - p$

এখন,  $p = - q$

বামপক্ষ= $\frac{1}{{(p+q+r)}^5}=\frac{1}{{(-q+q+r)}^5}=\frac{1}{r^5}$

ডানপক্ষ$=\frac{1}{p^5}+\frac{1}{q^5}+\frac{1}{r^5}=\frac{1}{{\left(-q\right)}^5}+\frac{1}{q^5}+\frac{1}{r^5}=-\frac{1}{q^5}+\frac{1}{q^5}+\frac{1}{r^5}=\frac{1}{r^5}$

সুতরাং $\frac{1}{{(p+q+r)}^5}=\frac{1}{p^5}+\frac{1}{q^5}+\frac{1}{r^5}$ (দেখানো হলো)