Question
Download Solution PDFयदि (a + b - c) = 20, तथा a2+ b2 + c2 = 152 है, तो a3 + b3 - c3 + 3abc का मान ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया:
(a + b - c) = 20;
a2+ b2 + c2 = 152
प्रयुक्त सूत्र:
a3 + b3 = (a + b)3 - 3ab × (a + b)
गणना:
इस समीकरण में 3 चर हैं और 2 समीकरण दिए गए हैं।
तो मान लीजिए c = 0
⇒ (a + b) = 20
⇒ a2 + b2 = 152
अब,
(a + b) = 20
दोनों पक्षों का वर्ग करना
⇒ 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
∴ सही उत्तर 560 है।
Alternate Method अवधारणा:
आवश्यक व्यंजक ज्ञात करने के लिए बीजीय सर्वसमिकाओं का उपयोग करें।
गणना
घनों के योग के लिए सर्वसमिका का उपयोग करें:
हम जानते हैं,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
यदि c ⇒ - c, तो
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)}
दिया गया:
a + b - c = 20
a2 + b2 + c2 = 152
ज्ञात करना ( ab + bc - ca ) :
⇒ (a + b - c)2 = a2 + b2 + c2 + 2ab - 2bc - 2ca
⇒ 202 = 152 + 2ab - 2bc - 2ca
⇒ 400 = 152 + 2(ab - bc - ca)
⇒ 248 = 2(ab - bc - ca)
⇒ ab - bc - ca = 124
अभिव्यक्ति में प्रतिस्थापित करें:
a3 + b3 - c3 + 3abc
⇒(a + b - c){(a2 + b2 + c2 - (ab - bc - ca)}
= 20 × (152 - 124)
= 20 × 28
= 560
∴ a3 + b3 - c3 + 3abc का मान 560 है।
Last updated on Jul 8, 2025
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