Question

Find the value -

${(\sqrt{3}+1)}^4+(\sqrt{3}-1{)}^4=?$

Answer 1

$(\sqrt{3}+1{)}^4+(\sqrt{3}-1{)}^4$

Using binomial expansion

$=\left[{}^4{C}_0{\left(\sqrt{3}\right)}^4{\left(1\right)}^0{+}^4{C}_1{\left(\sqrt{3}\right)}^3{\left(1\right)}^1{+}^4{C}_2{\left(\sqrt{3}\right)}^2{\left(1\right)}^2{+}^4{C}_3{\left(\sqrt{3}\right)}^1{\left(1\right)}^3{+}^4{C}_4{\left(\sqrt{3}\right)}^0{\left(1\right)}^4\right]+\left[{}^4{C}_0{\left(\sqrt{3}\right)}^4{\left(-1\right)}^0{+}^4{C}_1{\left(\sqrt{3}\right)}^3{\left(-1\right)}^1{+}^4{C}_2{\left(\sqrt{3}\right)}^2{\left(-1\right)}^2{+}^4{C}_3{\left(\sqrt{3}\right)}^1{\left(-1\right)}^3{+}^4{C}_4{\left(\sqrt{3}\right)}^0{\left(-1\right)}^4\right]$

$={\left(\sqrt{3}\right)}^4+4{\left(\sqrt{3}\right)}^3+6{\left(\sqrt{3}\right)}^2+4\left(\sqrt{3}\right)+1+{\left(\sqrt{3}\right)}^4-4{\left(\sqrt{3}\right)}^3+6{\left(\sqrt{3}\right)}^2-4\left(\sqrt{3}\right)+1$

$=2(\sqrt{3}{)}^4+12(\sqrt{3}{)}^2+2$

$=56$