Answer the following mathematical problems:

$Here, L.C.M. of 6, 7 and 8$

$$\begin{gathered} 6, 7, 83, 7, 42 \\ \therefore L.C.M.=2\times 3\times 7\times 4=168 \end{gathered}$$

$$\begin{gathered} We have 264 oranges. \\ \therefore We need = 168 \times 2 = 336 oranges. \end{gathered}$$

$\therefore We need additional = 336-264 = 72$

$\therefore Required number of oranges is 72. \left(answer\right)$

$Here, principal amount, D= 15Tk .; interest rate, r = 5\%$

$Number of years, n = 3 number of months, m = 4 [Quarterly means 4 times a year]$

$$\begin{gathered} Now, we know required amount, ‍A=p{(1+\frac{r}{100\times m})}^{mn}=15000{(1+\frac{5}{100\times 4})}^{4\times 3} \\ =15000{(1+\frac{1}{80})}^{12} \\ =15000{\left(\frac{81}{80}\right)}^{12} \\ =15000{(1.0125)}^{12} \\ =17411.317 Tk. \end{gathered}$$

$\therefore Required amount is 17,411.317 Tk.$

$$\begin{gathered} Given that, x^4+\frac{1}{x^4}=119 \\ ={\left(x^2\right)}^2+{\left(\frac{1}{x^2}\right)}^2=119 \\ ={(x^2+\frac{1}{x^2})}^2-2\times x^2\times \frac{1}{x^2}=119 \\ ={(x^2+\frac{1}{x^2})}^2-2=119 ={(x^2+\frac{1}{x^2})}^2=121 =x^2+\frac{1}{x^2}=\sqrt{121}=\sqrt{{\left(11\right)}^2}=11 \\ ={(x-\frac{1}{x})}^2+2\times x\times \frac{1}{x}=11 \\ ={(x-\frac{1}{x})}^2+2=11 \\ ={(x-\frac{1}{x})}^2=11-2=9 \\ \therefore x-\frac{1}{x}=3 \end{gathered}$$

$$\begin{gathered} Now, L.H.S. = x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3\times x\times \frac{1}{x}(x-\frac{1}{x}) \\ =3^3+3\times 3=27+9=36 \\ \therefore L.H.S.=R.H.S.\left(Proved\right) \end{gathered}$$

$Let, ABC is the right angle triangle$

$Here, Hypotenuse AB = (2x - 6)$

$$\begin{gathered} One arm, AC = x \\ And another arm, BC = (x + 2) \\ \therefore According to Pethagorean Theorem, \\ {(2x-6)}^2={(x+2)}^2+x^2 \end{gathered}$$

$={\left(2x\right)}^2-2\times 2x\times 6+6^2=x^2+2\times x\times 2+2^2+x^2$

$$\begin{gathered} =4x^2-24x+36=2x^2+4x+4 \\ =2x^2-28x+32=0 \\ =x^2-14x+16=0 \end{gathered}$$

$$\begin{gathered} Now, comparing, x^2-14x+16=0 with a{x}^2+bx+c=0, \\ We get, x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-14)\pm \sqrt{{(-14)}^2-4\times 1\times 16}}{2\times 1}=\frac{14\pm \sqrt{196-64}}{2}=\frac{14\pm \sqrt{132}}{2}=\frac{14\pm \sqrt{4\times 33}}{2} \end{gathered}$$

$$\begin{gathered} =\frac{14\pm 2\sqrt{33}}{2}=\frac{2(7\pm \sqrt{33})}{2}=7\pm \sqrt{33} \\ \therefore x=(7+\sqrt{33}) or (7-\sqrt{33}) \end{gathered}$$