ラマヌジャンの公式集：初等的な恒等式

$$\begin{array}{rl}\sqrt{2(\sqrt{a^2+b^2}-a)(\sqrt{a^2+b^2}-b)}& =a+b-\sqrt{a^2+b^2}\\ \sqrt[3]{3(\sqrt[3]{a^3+b^3}-a)(\sqrt[3]{a^3+b^3}-b)}& =\sqrt[3]{(a+b{)}^2}-\sqrt[3]{a^2-ab+b^2}\end{array}$$

$$B{k}^4-4A{k}^3-8Bkp-4Ap=0$$
において
$$\sqrt{A+B\sqrt[3]{p}}=\sqrt{\frac{B}{p+k^3}}(\frac{1}{2}{k}^2+k\sqrt[3]{p}-\sqrt[3]{p^2})$$

$$\sqrt{m\sqrt[3]{4m-8n}+n\sqrt[3]{4m+n}}=\frac{1}{3}(\sqrt[3]{(4m+n{)}^2}+\sqrt[3]{4(m-2n)(4m+n)}-\sqrt[3]{2(m-2n{)}^2})$$

$$\begin{array}{rl}& \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3m{n}^2+n^3-m^3}\\ =& \sqrt[3]{\frac{(m-n)(m+2n{)}^2}{9}}-\sqrt[3]{\frac{(2m+n)(m-n{)}^2}{9}}+\sqrt[3]{\frac{(m+2n)(2m+n{)}^2}{9}}\end{array}$$

$$\begin{array}{rl}\sqrt[3]{\frac{3}{2}(\sqrt[3]{9}-1)}& =\sqrt[3]{\cos {40}^{\circ }}+\sqrt[3]{\cos {80}^{\circ }}-\sqrt[3]{\cos {20}^{\circ }}\\ \sqrt[3]{6(\sqrt[3]{9}-1)}& =\frac{1}{\sqrt[3]{\cos {40}^{\circ }}}+\frac{1}{\sqrt[3]{\cos {80}^{\circ }}}-\frac{1}{\sqrt[3]{\cos {20}^{\circ }}}\end{array}$$

$$p^3+q^3+r^3=s^3$$
において
$$m=\pm (s+q)\sqrt{\frac{s-q}{r+p}},n=\pm (r-p)\sqrt{\frac{r+p}{s-q}}$$
とおくと
$$(p{a}^2+mab-r{b}^2{)}^3+(q{a}^2-nab+s{b}^2{)}^3+(r{a}^2-mab-p{b}^2{)}^3=(s{a}^2-nab+q{b}^2{)}^3$$

$$(a+b{)}^2=x+n{a}^2=y+nab=z+n{b}^2$$
において
$$x^2+(n-2)xz+z^2=n{y}^2$$

$$(a+b{)}^2=p+3a^2=q+3ab=r+3b^2$$
つまり
において
$$n(mp+nq{)}^3+m(mq+nr{)}^3=m(np+mq{)}^3+n(nq+mr{)}^3$$

$$(3a^2+5ab-5b^2{)}^3+(4a^2-4ab+6b^2{)}^3+(5a^2-5ab-3b^2{)}^3=(6a^2-4ab+4b^2{)}^3$$

$$\begin{array}{rl}(8s^2+40st-24t^2{)}^4+(6s^2-44st-18t^2{)}^4+(14s^2-4st-42t^2{)}^4& \\ +(9s^2+27t^2{)}^4+(4s^2+12t^2{)}^4& =(15s^2+45t^2{)}^4\\ (4m^2-12n^2{)}^4+(3m^2+9n^2{)}^4+(2m^2-12mn-6n^2{)}^4& \\ +(4m^2+12n^2{)}^4+(2m^2+12mn-6n^2{)}^4& =(5m^2+15n^2{)}^4\end{array}$$

　$a+b+c=0$において
$$\begin{array}{rl}2(ab+bc+ca{)}^2& =a^4+b^4+c^4\\ 2(ab+bc+ca{)}^4& =a^4(b-c{)}^4+b^4(c-a{)}^4+c^4(a-b{)}^4\\ 2(ab+bc+ca{)}^6& =(a^{2}b+b^{2}c+c^{2}a{)}^4+(a{b}^2+b{c}^2+c{a}^2{)}^4+(3abc{)}^4\\ 2(ab+bc+ca{)}^8& =(a^3+2abc{)}^4(b-c{)}^4+(b^3+2abc{)}^4(c-a{)}^4+(c^3+2abc{)}^4(a-b{)}^4\end{array}$$

　$ad=bc$および$n=2,4$において
$$(a+b+c{)}^n+(b+c+d{)}^n+(a-d{)}^n=(c+d+a{)}^n+(d+a+b{)}^n+(b-c{)}^n$$

　$ad=bc$において
$$\begin{array}{rl}& 64((a+b+c{)}^6+(b+c+d{)}^6-(c+d+a{)}^6-(d+a+b{)}^6+(a-d{)}^6-(b-c{)}^6)\\ & \times ((a+b+c{)}^{10}+(b+c+d{)}^{10}-(c+d+a{)}^{10}-(d+a+b{)}^{10}+(a-d{)}^{10}-(b-c{)}^{10})\\ =& 45((a+b+c{)}^8+(b+c+d{)}^8-(c+d+a{)}^8-(d+a+b{)}^8+(a-d{)}^8-(b-c{)}^8{)}^2\end{array}$$

　$3k=a^2+ab+b^2$において
$$(a{x}^3+k{)}^3-(b{x}^3+k{)}^3=k((x^4+ax{)}^3-(x^4+bx))$$

$$(x^4+1{)}^4+4{\left(\frac{x^5-5x}{2}\right)}^4+5(x^4-2{)}^4=3^4+4{\left(\frac{x^5+x}{2}\right)}^4$$

$$(4x^4+1{)}^4+(4x^5-5x{)}^4+5(4x^4-2{)}^4=3^4+(4x^5+x{)}^4$$

$$(4x^4+1{)}^4+(4x^5-5x{)}^4+(6x^4-3{)}^4=3^4+(4x^5+x{)}^4+(2x^4-1{)}^4$$

　${\alpha }^2+\alpha \beta +{\beta }^2=3\lambda {\gamma }^2$において
$$(\alpha +{\lambda }^2\gamma {)}^3+(\lambda \beta +\gamma {)}^3=(\lambda \alpha +\gamma {)}^3+(\beta +{\lambda }^2\gamma {)}^3$$

$$\sum _{n=0}^{\mathrm{\infty }}\lfloor \frac{n}{p}\rfloor x^n=\frac{x^p}{(1-x)(1-x^p)}$$

$$\lfloor \frac{n+4}{6}\rfloor -\lfloor \frac{n+3}{6}\rfloor +\lfloor \frac{n+2}{6}\rfloor =\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n}{3}\rfloor$$

$$\lfloor \sqrt{n+1}+\sqrt{n}\rfloor =\lfloor \sqrt{4n+2}\rfloor$$

$$\lfloor \frac{1}{2}+\sqrt{n+\frac{1}{2}}\rfloor =\lfloor \frac{1}{2}+\sqrt{n+\frac{1}{4}}\rfloor$$