$4x–3=5$ হলে, প্রমাণ কর যে, $64x^3–27-180x=125$

$(3x+y{)}^3$

$=(3x{)}^3+3\times (3x{)}^2\times y+3\times \left(3x\right)\times (y{)}^2+(y{)}^3$

$=27x^3+3\times 9x^2\times xy+3\times 3x\times y^2+y^3$

$=27x^3+27x^{2}y+9x{y}^2+y^3$ (Ans.)

$(x^2+y{)}^3=(x^2{)}^3+3(x^2{)}^2\times y+3\times x^2\times (y{)}^2+(y{)}^3$

$=x^6+3\times x^4\times x\times y+3\times x^2\times x^2\times y^2+y^3$

$=x^6+3x^{4}y+3x^2{y}^2+y^3$ (Ans.)

$(5p+2q{)}^3$

$=(5p{)}^3+3\times (5p{)}^2\times 2q+3\times 5p\times (2q{)}^2+(2q{)}^3$

$=125p^3+3\times 25p^2\times 2q+3\times 5p\times 4q^2+8q^3$

$=125p^3+150p^{2}q+60p{q}^2+8q^3\mathbf{(}\mathit{A}\mathit{n}\mathit{s}\mathbf{.}\mathbf{)}$

$(a^{2}b+c^{2}d{)}^3$

$=(a^{2}b{)}^3+3\times (a^{2}b{)}^2\times \left(c^{2}d\right)+3\times \left(a^{2}b\right)\times (c^{2}d{)}^2+(c^{2}d{)}^3$

$=a^6{b}^3+3a^4{b}^2{c}^{2}d+3a^{2}b{c}^4{d}^2+c^6{d}^3$ (Ans.)

$(6p-7{)}^3$

$=(6p{)}^3-3\times (6p{)}^2\times 7+3\times \left(6p\right)\times (7{)}^2-(7{)}^3$

$=216p^3-3\times 36p^2\times 7+3\times 6p\times 49-343$

$=216p^3-756p^2+882p-343$ (Ans.)

$(ax-by{)}^3$

$=(ax{)}^3-3\times (ax{)}^2\times by+3\times ax\times (by{)}^2-(by{)}^3$

$=a^3{x}^3-3\times a^2{x}^2\times by+3\times ax\times b^2{y}^2-b^3{y}^3$

$=a^3{x}^3-3a^2{x}^2 by + 3ax{b}^2{y}^2-b^3{y}^3$ (Ans.)