Question

verify a³+b³ =(a+b)(a²+b²-ab)​

Answer 1

For Fig. 10.1, volume of cuboid = (a-b) x a x a = (a-b)a²

For Fig. 10.2, volume of cuboid = (a-b) x a x b = (a-b)ab

For Fig. 10.3, volume of cuboid = (a – b) x b x b = (a – b)b²

For Fig. 10.4, volume of cube =b³

For Fig. 10.5, volume of cube = Sum of volume of all cubes and cuboids

= (a – b)a² + (a – b)ab + (a – b)b² + b³ …..(i)

The cube obtained in Fig. 10.5 has its each side a.

Its volume = (side)³ = a³ …..(ii)

From Eqs. (i) and (ii), we get

a³ = (a – b)a² + (a – b)ab + (a – b)b² + b³ …..(iii)

For Fig. 10.6, volume of solid obtained = a³ – b³

= (a – b)a² + (a – b)ab + (a – b)b² + b³ – b³ [from Eq.(iii)]

= (a – b)a² + (a – b)ab + (a – b)b² = (a-b) (a² +ab + b²)

Therefore, a³-b³ = (a-b) (a²+ab+b²)

Here, volume is in cubic units.

Observation

On actual measurement, we get

a = ……. , b = ……. ,

So, a² =…….. , b² = ……. ,

(a- b) = ……. , ab = ……. ,

a³ =…….. , b³ = ……. ,

Hence, a³-b³ = (a-b) (a²+ab+b²).

Result

The algebraic identity a³-b³ = (a-b) (a²+ab+b²) has been verified.

Answer 2

${a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} ) \\ \\ rhs \\ \\ = a( {a}^{2} - ab + {b}^{2}) + b({a}^{2} - ab + {b}^{2}) \\ \\ = {a}^{3} - {a}^{2} b + a {b}^{2} + {a}^{2} b - a {b}^{2} + {b}^{3} \\ \\ = {a}^{3} + {b}^{3}$