Which two numbers, from amongst the given options, should be interchanged to make the given equation correct?

[(\(\sqrt{256}\)) ÷ 4]³ + (6)² - (12 × 8) + 66 + (36 ÷ 12) = 65

Given statement : [(\(\sqrt{256}\)) ÷ 4]3 + (6)2 - (12 × 8) + 66 + (36 ÷ 12) = 65

According to the question, after interchanging the given numbers :

So,

• Option - (1) : 6 and 8

[(\(\sqrt{256}\)) ÷ 4]3 + (8)2 - (12 × 6) + 66 + (36 ÷ 12) = 65

[16 ÷ 4]3 + 64 - 72 + 66 + 3 = 65

[4]3 + 64 - 72 + 66 + 3 = 65

64 + 64 + 66 + 3 - 72 = 65

197 - 72 = 65

125 ≠ 65 (LHS ≠ RHS)

• Option - (2) : 66 and 65

[(\(\sqrt{256}\)) ÷ 4]3 + (6)2 - (12 × 8) + 65 + (36 ÷ 12) = 66

[16 ÷ 4]3 + 36 - 96 + 65 + 3 = 66

[4]3 + 36 - 96 + 65 + 3 = 66

64 + 36 + 65 + 3 - 96 = 66

168 - 96 = 66

72 ≠ 65 (LHS ≠ RHS)

• Option - (3) : 4 and 8

[(\(\sqrt{256}\)) ÷ 8]3 + (6)2 - (12 × 4) + 66 + (36 ÷ 12) = 65

[16 ÷ 8]3 + 36 - 48 + 66 + 3 = 65

[2]3 + 36 - 48 + 66 + 3 = 65

8 + 36 + 66 + 3 - 48 = 65

113 - 48 = 65

65 = 65 (LHS = RHS)

• Option - (4) : 66 and 36

[(\(\sqrt{256}\)) ÷ 4]3 + (6)2 - (12 × 8) + 36 + (66 ÷ 12) = 65

[16 ÷ 4]3 + 36 - 96 + 36 + 5.5 = 65

[4]3 + 36 - 96 + 36 + 5.5 = 65

64 + 36 + 36 + 5.5 - 96 = 65

141.5 - 96 = 65

45.5 ≠ 65 (LHS ≠ RHS)

Hence, "Option - (3)" is the correct answer.