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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 1 Evaluate: (𝑖^18+(1/i)^25 )^3 (𝑖^18+(1/𝑖)^25 )^3 = (𝑖^18+ 1/(π’Š)^πŸπŸ“ )^3 = (𝑖^18+ 1/(π’Š Γ— π’Š^πŸπŸ’ ))^3 = ((π’Š^𝟐 )^πŸ—+1/(𝑖 Γ— (π’Š^𝟐 )^𝟏𝟐 ))^3 Putting i2 = βˆ’πŸ = ((βˆ’πŸ)^πŸ—+1/〖𝑖 Γ— (βˆ’πŸ)γ€—^12 )^3 = (βˆ’πŸ+1/(𝑖 Γ— 𝟏))^3 = (βˆ’1+1/𝑖)^3 Removing π’Š from the denominator = (βˆ’1+1/π‘–Γ—π’Š/π’Š)^3 = (βˆ’1+𝑖/π’Š^𝟐 )^3 = (βˆ’1+𝑖/((βˆ’πŸ)))^3 = (βˆ’πŸ – π’Š )πŸ‘ = (βˆ’1(1+ 𝑖 ))3 = (βˆ’πŸ)πŸ‘ (𝟏 + π’Š )πŸ‘ = (βˆ’1)(1 + 𝑖 )3 = βˆ’(𝟏 + π’Š )πŸ‘ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = βˆ’(13 + 𝑖3 + 3 Γ— 1 Γ— 𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ‘ +3𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ Γ— π’Š +3𝑖 (1 + 𝑖)) Putting i2 = βˆ’πŸ = βˆ’(1 +(βˆ’πŸ) Γ— 𝑖 +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖+3𝑖 Γ— 𝑖) = βˆ’(1+2𝑖+3π’Š^𝟐 ) Putting i2 = βˆ’πŸ = βˆ’(1+2𝑖+3 Γ— βˆ’πŸ) = βˆ’(1+2π‘–βˆ’3) = βˆ’(2π‘–βˆ’2) = βˆ’2𝑖+2 = πŸβˆ’πŸπ’Š

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.