Question

factorise a^3-b^3+1+3ab

Answer 1

a^3-b^3+1+3ab

=> a^3+(-b)^3+1^3-3(1*a*(b))

=> {a+(-b)+1){a^2+(-b)^2+1^2-a(-b)-(-b) 1-1a}

=> (a-b+1)(a^2+b^2+1+ab+b-a)

Hope this helps!

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

TO FACTORISE :-

$\huge{ \sf{ \pink{a {}^{3} - b {}^{3} + 1 + 3ab}}}$

$\: \: \: \: \: \: \tt{ \red{a {}^{3} - b {}^{3} + 1 + 3ab }}\\ \\ \tt{ \blue{= (a) {}^{3} + ( - b) {}^{3} + (1) {}^{3} - 3 \times a \times ( - b) \times 1 }}\\ \\ \tt{ \green{ = \{a + ( - b )+ 1 \} \{ (a) {}^{2} + ( - b) {}^{2} +( 1) {}^{2} - a \times ( - b) - }} \\\tt{ \green{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (- b) \times 1 - 1 \times a \}}} \\ \\ \tt{ \blue{= (a - b + 1)(a {}^{2} + b {}^{2} + 1 + ab + b - a)}} \\ \\ \tt{ \red{= (a - b + 1)(a {}^{2} + b {}^{2} + 1 - a + b+ ab)}}$