(i) ∛100 × ∛270
As we know ∛a × ∛b = ∛(a×b)
Now,
∛100 × ∛270 = ∛(100×270)
= ∛ (2×2×5×5) × (2×3×3×3×5)
= ∛(23×33×53)
= 2×3×5
= 30
(ii) ∛121 × ∛297
As we know ∛a × ∛b = ∛(a×b)
Now,
∛121 × ∛297 = ∛ (121×297)
= ∛ (11×11) × (3×3×3×11)
= ∛(113×33)
= 11×3
= 33
