যদি (a + b - c) = 20 এবং a2 + b2 + c2 = 152 হয়, তাহলে a3 + b3 - c3 + 3abc এর মান নির্ণয় কর।

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SSC CGL 2023 Tier-I Official Paper (Held On: 25 Jul 2023 Shift 4)
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  1. 480
  2. 720
  3. 640
  4. 560

Answer (Detailed Solution Below)

Option 4 : 560
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Free
PYST 1: SSC CGL - General Awareness (Held On : 20 April 2022 Shift 2)
25 Qs. 50 Marks 10 Mins

Detailed Solution

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প্রদত্ত:

(a + b - c) = 20;

a2 + b2 + c2 = 152

সূত্র ব্যবহৃত:

a3 + b3 = (a + b)3 - 3ab × (a + b)

গণনা:

আমাদের এই সমীকরণে 3টি ভেরিয়েবল রয়েছে এবং 2টি সমীকরণ দেওয়া হয়েছে।

তারপর c = 0 দিন

(a + b) = 20

a2 + b2 = 152

এখন,

(a + b) = 20

উভয় পক্ষের স্কোয়ারিং

⇒ a2 + b2 + 2ab = 400

⇒ 152 + 2ab = 400

⇒ 2ab = 400 - 152 = 248

⇒ ab = 248/2 = 124

a3 + b3 = (a + b)3 - 3ab × (a + b)

⇒ (20)3 - 3 × 124 × 20

⇒ 8000 - 7440 = 560

∴ সঠিক উত্তর হল 560।

Alternate Method 
ধারণা:

প্রয়োজনীয় রাশি খুঁজতে বীজগণিতীয় পরিচয় ব্যবহার করুন।

হিসাব

ঘনক্ষেত্রের যোগফলের জন্য পরিচয় ব্যবহার করুন:

আমরা জানি,

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

যদি c ⇒ - c, তাহলে

a3 + b3 - c3 + 3abc = (a + b - c){(a2 + b2 + c2 - ab - b(-c) - (-c)a)}

⇒ a3 + b3 - c3 + 3abc = (a + b - c)(a2 + b2 + c2 - ab + bc + ca)

⇒ a3 + b3 - c3 + 3abc = (a + b - c){(a2 + b2 + c2 - (ab - bc - ca)} 

প্রদত্ত:

a + b - c = 20

a2 + b2 + c2 = 152

খুঁজুন

( ab + bc - ca):

⇒ (a + b - c)2 = a2 + b2 + c+ 2ab - 2bc - 2ca

⇒ 202 = 152 + 2ab - 2bc - 2ca

⇒ 400 = 152 + 2(ab - bc - ca) 

⇒ 248 = 2(ab - bc - ca) 

⇒ ab - bc - ca = 124 

অভিব্যক্তিতে বিকল্প:

a3 + b3 - c3 + 3abc

⇒(a + b - c){(a2 + b2 + c2 - (ab - bc - ca)} 

= 20 ×  (152 - 124)

= 20 × 28 

= 560 

∴ a3 + b3 - c3 + 3abc এর মান হল 560।

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